# CBSE Class 12 Maths Revision Notes Chapter 7

## Class 12 Mathematics Chapter 7 Notes

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## NCERT Class 12 Mathematics Chapter 7 Notes: Main Topics

The main topics covered in this chapter are as follows:

• Introduction
• Integration
• Indefinite Integrals
• Application of Integrals
• Integral Calculus

An overview of Class 12 Mathematics cChapter 7 notes is given below.

### INTRODUCTION:

The Class 12 Mathematics chapter 7 notes include various complex concepts related to Integration and application of Integrals.

Integration is the inverse of differentiation. Integration is defined as the process of determining a function, say F(x), whose differential coefficient is known.
Therefore, Let the differential coefficient of F(x) be f(x).
ddx[F(x)]=f(x), we say that F(x) is integral or antiderivative of f(x).
Integration is denoted by f(x) dx= F(x) where,
f(x) is a function f of variable x is known as the integral and f(x) dx is known as the element of integration.

#### INDEFINITE INTEGRAL

If ddx[F(x)]=f(x) then arbitrary constant C, f(x) dx= F(x) + C.
This shows that F(x) and F(x) + C are integrals of the same function f(x). If the value of C varies, we get different values of integrals of f(x). Therefore, the integral of the function f(x) is not definite. By virtue of this property, we can say that F(x) is the indefinite integral of f(x).

Properties:

1. [f(x) + g(x) dx] = f(x) dx + g(x) dx
2. ddx[F(x)]=f(x)
3. k.f(x) = k. f(x)dx
4. If f1(x), f2(x), f3(x)…… fn(x) are functions and k1, k2,, k3…..kn are real numbers. Then
[r1f1(x) r2f2(x)r3 f3(x)…… rnfn(x)] dx = r1f1(x) dx r2f2(x) dx … rnfn(x) dx

#### Standard Formulas:

1. xndx = xn+1n+1 + C, n -1
2. 1xdx = log x+C
3. exdx = ex + C
4. axdx = axlogea+C
5. sin x dx =-cos x +C
6. cos x dx=sin x+C
7. sec2x dx=tan x+C
8. sec x. tan x dx=sec x+C
9. cosec x. cot x dx=-cosec x+C
10. tan x dx= log|cosx|+ C = log|secx|+ C
11. cot x dx = log|sinx|+ C
12. sec x dx = log|secx + tanx|+ C
13. cosec x dx = log|cosecx − cotx|+ C
14. 11-x2dx = sin-1x +C
15. 11+x2dx = tan-1x +C
16. 1xx2-1dx = sec-1x +C

### Geometric Interpretation:

If d/dx[F(x)]=f(x), then f(x) dx= F(x) + C. If the value of C varies, we get different values of integrals of f(x), differing by a constant. The graph of the function depicts an infinite family of curves and they have the same slope F’(x) = f(x).

### METHODS OF INTEGRATION:

#### 1. Substitution Method:

By using the substitution method, the variable x in f(x) dx changes into another variable, t. Therefore, the integrand f(x) changes into F(t), which is known to be the algebraic sum of standard integrals. There is no formula to determine a suitable substitute.

a. If given integrand is of the form f’(ax + b),
We substitute ax+b=t. Therefore, dx= 1a dt
Integrating on both sides, we get
f′(ax+b)dx = f'(t)1a dt = f(t)a =f(ax + b)a+C

b. If given integrand is of the form xn-1f’(xn),
We substitute xn=t. Therefore, n.xn-1dx= dt
Integrating on both sides, we get
xn-1f’(xn)dx = f'(t)dtn = 1n =f(ax + b)a+C
Therefore, f'(t) dt = 1n f(t) = 1n f(xn) + C

c. If given integrand is of the form [f(x)n]f’(x),
We substitute f(x)=t. Therefore, f’(x) dx= dt

d. If given integrand is of the form f’(x)f(x),
We substitute f(x)=t. Therefore, f’(x) dx= dt
Integrating on both sides, we get
f’(x)f(x) dx = dtt = log t=log f(x)+C

#### Some Special Integrals:

dxx2+a2=1a tan-1xa+C
dxx2-a2=12alogx-ax+a+C
dxa2-x2=12aloga+xa-x+C
1a2-x2dx = sin-1xa +C
1×2+a2dx = log x+x2+a2 + C
1×2-a2dx = log x+x2-a2 + C
a2-x2 dx = x2a2-x2+ a22 sin-1xa +C
x2+a2 dx = x2x2+a2+ a22 log x+x2+a2 + C
x2-a2 dx = x2x2-a2+ a22 log x+x2-a2 + C

Rules to solve integrations of the following forms through Integral Substitution
∫f(a2-x2)dx, substitute x=a.sin⁡ θ or x=a.cos θ
∫f(a2+x2)dx, substitute x=a.tan θ or x=a.cot θ
∫f(x2-a2)dx, substitute x=a.sec θ or x=a.cosec θ
f(a+xa-x) dx or f(a-xa+x) dx, substitute x=a.cos 2θ

#### Integrals of the form 1:

dxax2+bx+c
dxax2+bx+c
ax2+bx+c dx

Rules to solve the above-mentioned integrals:
Step 1: Make the coefficient of x2 =1. Take the coefficient of x2 common from the quadratic equation.
Step 2: ax2+bx+c in the form of a[(x + b2a)2] – b2-4ac2a
Step 3: Use one of the special integrals to transform the integrand.
Step 4: Integrate the function.

#### Integrals of the form 2:

px+qax2+bx+cdx
px+qax2+bx+cdx
(px+q)ax2+bx+c dx
To solve these integrals, we carry out the following steps:
Substitute px+q as λ (2ax+b) + μ or px+q= λ+ μ
On comparing, we get
p = 2aλ and q = bλ+ μ, which is equal to
λ=p2a and μ = q – bλ μ = q – bp2a
Using these, we transform the integrand and thus integrate the function.

#### Integrals of form 3:

P(x) ax2+bx+cdx, where P(x) is a polynomial of degree ⩾ 2.
This equation becomes (a0+a1x+a2x2….. + an-1xn-1)ax2+bx+c + k dx ax2+bx+c.
The value of the constants is found by separating the relation and comparing the coefficients of the different powers of x on both sides.

#### Integrals of the form 4:

x2+ 1×4+ kx2+ 1dx or x2- 1×4+ kx2+ 1dx , where k is a constant
Rules to solve the above-mentioned integrals:
Step 1: The numerator and denominator are divided by x2
Step 2: Substitute z = x + 1x or z = x – 1x
Step 3: Integrate the function with respect to z
Step 4: Express the answer in terms of variable x

#### Integrals of the form 5:

dxPQ, Let P = ax+b and Q = cx+d (linear or quadratic equations)
Substitute cx + d = z2
Integrate the function with respect to z and express the solution in terms of variable x

#### Integrals of the form 6:

dxa+b cos x
dxa+b sin x
dxa+b cos x + c sin x

Rules:
Step 1: Substitute cos x = 1 – tan2x21 + tan2x2 and sin x = 2 tanx21 + tan2x2
Step 2: Put tanx2 = z
Step 3: Integrate the function

#### Integrals of the form 7:

dxa+b cos2x
dxa+b sin2 x
dxa cos2x+b sin x cos x + c sin2x

Rules:
Step 1: Divide entire function by cos2x
Step 2: Substitute sec x = 1 + tan2x
Step 3: Substitute z = tan x dz = sec2x dx
Step 4: Integrate the function

#### Integrals of form 8:

a cos x + b sin xc cos x + d sin x
Rules:
Step 1: Substitute a.cosx+b.sinx= λ(c.cosx+d.sinx) + μ(−c.sinx+d.cosx)
Step 2: Find values of λ and μ by equating sin x and cos x
Step 3: DIvide the Function into two parts and substitute the value of λ and μ

#### Integrals of the form 9:

a + b cos x + c sin xd + e cos x + f sin x

Rules:
Step 1: Substitute a + b.cosx + c.sinx= l (e+f.cosx+g.sinx) +m (−f.sinx + g.cosx) + n
Step 2: Find values of l, m and n by equating sin x and cos x
Step 3: Divide the Function into three parts and substitute the value of l, m and n.

Refer to the Extramarks Class 12 chapter 7 Mathematics notes to practice additional problems on the above-mentioned concepts.

### METHOD OF PARTIAL FRACTIONS:

Integrals of the form p(x)q(x)dx are integrated with the help of partial fractions.
Firstly, check the degree of p(x) and q(x)

Substitute p(x)q(x) = r(x) + f(x)q(x) where degree of p(x) degree of q(x) > degree of f(x)
Case 1: Denominator has non-repeated linear components
q(x) = (x−1)(x−2)…(x−n)
Therefore, f(x)q(x) = A1 (x−1)+A2 (x−2)+…….+An (x−n)
Extract LCM and find the value of A1, A2, …… , An

Case 2: Denominator has repeated as well as non-repeated linear components.
q(x) = (x−1)2(x−3)…(x−n)
Therefore, f(x)q(x) = A1 (x−1)+A2 (x−1)2+…….+An (x−n)
Extract LCM and find the value of A1, A2, …… , An

Case 3: Denominator has a non-repeated quadratic component which is not further factorisable.
q(x) = (ax2+bx+c)(x−3)(x−4)…(x−n)
Therefore, f(x)q(x) = A1x+A2 (ax2+bx+c)+A3 (x−3)+A4 (x−4)…….+An (x−n)
Extract LCM and find the value of A1, A2, …… , An

Case 4: Denominator has a repeating quadratic component.
q(x) = (ax2+bx+c)2(x−5)(x−6)…(x−n)
Therefore, f(x)q(x) = A1x+A2 (ax2+bx+c)+A3x+A4 (ax2+bx+c)2+A5 (x−5)…….+An (x−n)
Extract LCM and find the value of A1, A2, …… , An

Case 5: The power x is even
Step 1: Substitute z= x2
Step 2: Resolve the functions in terms of z into partial fractions
Step 3: Substitute z= x2 again and carry out Integration.

### METHOD OF INTEGRATION BY PARTS

Let u and v be two functions, then integration of the product of u.v is given as
u.v dx= uv dx-(dudx. v.dx) dx
Case 1: Integrals of the form f(x).xn dx
Take xn = u i.e., the first function and f(x) as v.

Case 2: Integrals of form 1.(log x)n dx
Take (log x)n = u i.e., the first function and 1 as v.

Case 3: If the two functions u and v are of different types, then the first function can be chosen as
I – Inverse Trigonometric function
L – Logarithmic function
A – Algebraic function
T− Trigonometric function
E− Exponential function.
The sequence is remembered using an abbreviation, i.e., ‘ILATE’.
Integrals of the form: ex [f(x) + f′(x)] dx
Rules:
Divide these into two different integrals
Use Integration by parts to integrate the first part only.

Integrals of the form:
After Integration, if the initial integrand is formed again, then we solve it with the help of the following steps:
Use the Integration by part technique twice.
Substitute the repeating integrand as equal to I.
Further, solve for I.

Integration of hyperbolic functions:
sin hx dx =-cos hx +C
cos hx dx=sin hx+C
sec2hx dx=tan hx+C
cosec2hx dx=-cot hx+C
sec hx. tan hx dx=sec hx+C
cosec hx. cot hx dx=-cosec hx+C

### Chapter 7 Mathematics Class 12 Notes: Exercises & Solutions

The Class 12 Mathematics Chapter 7 notes include various questions for students to practice. The detailed and well-structured information helps students to get in-depth knowledge of Integration and, thus, apply that knowledge to solve the sums. Students will learn the rules to integrate a function of different forms. The Class 12 Mathematics Chapter 7 notes are simple to comprehend for students who desire to ace their examinations. It includes solutions to all exercise questions, miscellaneous problems as well as some CBSE extra questions. Students are recommended to solve the CBSE sample papers for more practice.

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Q.1

$\mathrm{Evaluate}\text{ }\int {\mathrm{sec}}^{4}\mathrm{x}\text{ }\mathrm{tanx}\text{ }\mathrm{dx}$

Ans

$\begin{array}{l}\int {\mathrm{sec}}^{4}\mathrm{x}\text{ }\mathrm{tanx}\text{ }\mathrm{dx}\\ =\text{ }\int {\mathrm{sec}}^{2}\mathrm{x}\text{ }{\mathrm{sec}}^{2}\mathrm{xtanx}\text{ }\mathrm{dx}\text{ }=\text{ }\int \left(1\text{ }+\text{ }{\mathrm{tan}}^{2}\mathrm{x}\right){\mathrm{sec}}^{4}\mathrm{x}\text{ }\mathrm{tanx}\text{ }\mathrm{dx}\\ \left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{tanx}\text{ }=\text{ }\mathrm{t}\\ {\mathrm{sec}}^{2}\mathrm{x}\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\end{array}\right]\\ =\text{ }\int \left(1+{\mathrm{t}}^{2}\right)\text{ }\mathrm{tdt}\text{ }=\text{ }\int \mathrm{t}+{\mathrm{t}}^{3}\text{ }\mathrm{dt}\\ =\text{ }\frac{{\mathrm{t}}^{2}}{2}\text{ }+\text{ }\frac{{\mathrm{t}}^{4}}{4}\text{ }+\text{ }\mathrm{C}\\ =\text{ }\frac{{\mathrm{tan}}^{2}\mathrm{x}}{2}\text{ }+\text{ }\frac{{\mathrm{tan}}^{4}\mathrm{x}}{4}\text{ }+\text{ }\mathrm{C}\end{array}$

Q.2

$\mathrm{Evaluate}\text{ }\int \frac{1}{1+\mathrm{sinx}}\text{ }\mathrm{dx}$

Ans

$\begin{array}{l}\int \frac{1}{1+\mathrm{sinx}}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{1}{1+\mathrm{sinx}}\text{ }×\text{ }\frac{1-\mathrm{sinx}}{1-\mathrm{sinx}}\mathrm{dx}\\ =\text{ }\int \frac{1-\mathrm{sinx}}{1-{\mathrm{sin}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{1-\mathrm{sinx}}{{\mathrm{cos}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\\ =\text{ }\int {\mathrm{sec}}^{2}\mathrm{x}-\text{ }\mathrm{secx}\text{ }\mathrm{tanx}\\ =\text{ }\mathrm{tanx}\text{ }-\text{ }\mathrm{sec}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{C}\end{array}$

Q.3

$\mathrm{Evaluate}\text{ }\int \frac{1}{1+\mathrm{tanx}}\text{ }\mathrm{dx}$

Ans

$\begin{array}{l}\int \frac{1}{1+\mathrm{tanx}}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{\mathrm{cosx}}{\mathrm{cosx}+\mathrm{sinx}}\text{ }\mathrm{dx}=\frac{1}{2}\int \frac{2\mathrm{cosx}}{\mathrm{cosx}+\mathrm{sinx}}\text{ }\mathrm{dx}\\ =\frac{1}{2}\int \frac{\mathrm{cosx}+\mathrm{cosx}}{\mathrm{cosx}+\mathrm{sinx}}\text{ }\mathrm{dx}\\ =\frac{1}{2}\int \frac{\mathrm{cosx}+\mathrm{sinx}\text{ }+\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\text{ }\mathrm{dx}\\ =\frac{1}{2}\int \frac{\mathrm{cosx}+\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\text{ }\mathrm{dx}\text{ }+\frac{1}{2}\int \frac{\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\text{ }\mathrm{dx}\\ =\frac{1}{2}\int 1\mathrm{dx}\text{ }+\frac{1}{2}\int \frac{\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\mathrm{dx}\\ =\frac{\mathrm{x}}{2}\text{ }+\frac{1}{2}\int \frac{1}{\mathrm{t}}\mathrm{dt}\text{ }\left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{cosx}\text{ }+\text{ }\mathrm{sin}\text{ }\mathrm{x}\text{ }=\text{ }\mathrm{t}\\ \left(-\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\text{ }\mathrm{dx}=\text{ }\mathrm{dt}\end{array}\right]\\ =\frac{\mathrm{x}}{2}\text{ }+\frac{1}{2}\text{ }\mathrm{log}\text{ }\left|\mathrm{t}\right|\text{ }+\text{ }\mathrm{C}\\ =\frac{\mathrm{x}}{2}\text{ }+\frac{1}{2}\text{ }\mathrm{log}\text{ }\left|\mathrm{cosx}\text{ }+\text{ }\mathrm{sin}\text{ }\mathrm{x}\right|\text{ }+\text{ }\mathrm{C}\end{array}$

Q.4

$\mathrm{Evaluate}\text{ }\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1+\mathrm{x}}\text{ }-\text{ }\sqrt{\mathrm{x}}}\text{ }\mathrm{dx}$

Ans

$\begin{array}{l}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1+\mathrm{x}}\text{ }-\text{ }\sqrt{\mathrm{x}}}\text{ }\mathrm{dx}\\ =\text{ }\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1+\mathrm{x}}\text{ }-\text{ }\sqrt{\mathrm{x}}}\text{ }×\text{ }\frac{\sqrt{1+\mathrm{x}}\text{ }+\text{ }\sqrt{\mathrm{x}}}{\sqrt{1+\mathrm{x}}\text{ }+\text{ }\sqrt{\mathrm{x}}}\mathrm{dx}\\ =\text{ }\underset{0}{\overset{1}{\int }}\frac{\sqrt{1+\mathrm{x}}\text{ }+\text{ }\sqrt{\mathrm{x}}}{1}\text{ }\mathrm{dx}\text{ }=\text{ }\underset{0}{\overset{1}{\int }}\sqrt{1+\mathrm{x}}\text{ }+\text{ }\sqrt{\mathrm{x}}\mathrm{dx}\\ =\text{ }{\left[\frac{2{\left(1+\mathrm{x}\right)}^{3/2}}{3}\right]}_{0}^{1}\text{ }+\text{ }{\left[\frac{2{\left(\mathrm{x}\right)}^{3/2}}{3}\right]}_{0}^{1}\\ =\text{ }\left[\frac{2{\left(1+1\right)}^{3/2}}{3}-\frac{2{\left(1+0\right)}^{3/2}}{3}\right]\text{ }+\text{ }\left[\frac{2{\left(1\right)}^{3/2}}{3}-\frac{2{\left(0\right)}^{3/2}}{3}\right]\\ =\text{ }\frac{2{\left(1+1\right)}^{3/2}}{3}\text{ }-\text{ }\frac{2}{3}\text{ }+\frac{2}{3}\\ =\text{ }\frac{2{\left(2\right)}^{3/2}}{3}\text{ }=\text{ }\frac{4\sqrt{2}}{3}\end{array}$

Q.5

$\mathrm{Evaluate}\text{ }\int {\mathrm{sec}}^{3}\mathrm{x}\text{ }\mathrm{dx}$

Ans

$\begin{array}{l}\int {\mathrm{sec}}^{3}\mathrm{x}\text{ }\mathrm{dx}\\ =\text{ }\int \mathrm{secx}\text{ }{\mathrm{sec}}^{2}\mathrm{xdx}\\ =\text{ }\int \sqrt{1+{\mathrm{tan}}^{2}\mathrm{x}}\text{ }{\mathrm{sec}}^{2}\mathrm{xdx}\\ \left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{tanx}\text{ }=\text{ }\mathrm{t}\\ {\mathrm{sec}}^{2}\text{ }\mathrm{xdx}\text{ }=\text{ }\mathrm{dt}\end{array}\right]\\ =\text{ }\int \sqrt{1+{\mathrm{t}}^{2}}\text{ }\mathrm{dt}\\ =\frac{\mathrm{t}}{2}\sqrt{1+{\mathrm{t}}^{2}}\text{ }+\frac{1}{2}\text{ }\mathrm{log}\text{ }\left|\mathrm{t}+\sqrt{1+{\mathrm{t}}^{2}}\right|\text{ }+\mathrm{C}\\ =\frac{\mathrm{tanx}.\mathrm{secx}}{2}\text{ }+\frac{1}{2}\text{ }\mathrm{log}\text{ }\left|\mathrm{tanx}+\mathrm{secx}\right|\text{ }+\mathrm{C}\end{array}$

Q.6

$\mathrm{Evaluate}\text{ }\int \sqrt{{\mathrm{x}}^{2}\text{ }-4\mathrm{x}\text{ }+2}\text{ }\mathrm{dx}$

Ans

$\begin{array}{l}\int \sqrt{{\mathrm{x}}^{2}\text{ }-4\mathrm{x}\text{ }+2}\text{ }\mathrm{dx}\\ =\int \sqrt{{\left(\mathrm{x}-2\right)}^{2}-2}\text{ }\mathrm{dx}\\ =\int \sqrt{{\left(\mathrm{x}-2\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}\text{ }\mathrm{dx}\\ =\frac{\mathrm{x}-2}{2}\text{ }\sqrt{{\mathrm{x}}^{2}\text{ }-4\mathrm{x}+2}-\text{ }\mathrm{log}\text{ }\left|\mathrm{x}-2+\sqrt{{\mathrm{x}}^{2}\text{ }-4\mathrm{x}+2}\right|\text{ }+\mathrm{C}\end{array}$

Q.7

$\mathrm{Evaluate}\text{ }\int \left(2\mathrm{x}+4\right)\text{ }\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}\text{ }+2}\mathrm{dx}$

Ans

$\begin{array}{l}\int \left(2\mathrm{x}+4\right)\text{ }\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}\text{ }+2}\mathrm{dx}\\ =\int \left(2\mathrm{x}+3+1\right)\text{ }\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}\text{ }+2}\mathrm{dx}\\ =\int \left(2\mathrm{x}+3\right)\text{ }\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}\text{ }+2}+1\text{ }\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}\text{ }+2}\mathrm{dx}\\ =\int \left(2\mathrm{x}+3\right)\text{ }\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}\text{ }+2}\mathrm{dx}+\int \sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}\text{ }+2}\text{ }\mathrm{dx}\\ =\int \sqrt{\mathrm{t}}\mathrm{dt}\text{ }+\int \sqrt{{\left(\mathrm{x}+3/2\right)}^{2}\text{ }-{\left(1/2\right)}^{2}}\text{ }\mathrm{dx}\text{ }\left[\begin{array}{l}\mathrm{Let}\text{ }{\mathrm{x}}^{2}+3\mathrm{x}+2\text{ }=\mathrm{t}\\ \left(2\mathrm{x}+3\right)\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\end{array}\right]\\ =\frac{2{\mathrm{t}}^{3/2}}{2}\text{ }+\text{ }\frac{\mathrm{x}+3/2}{2}\sqrt{{\mathrm{x}}^{2}+3\mathrm{x}+2}-\text{ }\frac{1}{8}\mathrm{log}\text{ }\left|\mathrm{x}+3/2+\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}+2}\right|\text{ }+\mathrm{C}\\ =\frac{2{\left({\mathrm{x}}^{2}+3\mathrm{x}+2\right)}^{3/2}}{3}\text{ }+\text{ }\frac{2\mathrm{x}+3}{4}\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}+2}-\text{ }\frac{1}{8}\mathrm{log}\text{ }\left|\mathrm{x}+3/2+\sqrt{{\mathrm{x}}^{2}\text{ }+3\mathrm{x}+2}\right|\text{ }+\mathrm{C}\end{array}$

Q.8

$\mathrm{Evaluate}\text{ }\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\frac{\mathrm{sinx}+\mathrm{cosx}}{9+16\mathrm{sin}2\mathrm{x}}\text{ }\mathrm{dx}$

Ans

$\begin{array}{l}\mathrm{I}=\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\frac{\mathrm{sinx}+\mathrm{cosx}}{9+16\mathrm{sin}2\mathrm{x}}\text{ }\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\frac{\mathrm{sinx}+\mathrm{cosx}}{9+16\left[1-{\left(\mathrm{sinx}-\mathrm{cosx}\right)}^{2}\right]}\text{ }\mathrm{dx}\\ \left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{sinx}-\mathrm{cosx}\text{ }=\mathrm{t}\\ \left(\mathrm{cosx}\text{ }+\text{ }\mathrm{sinx}\right)\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\\ \mathrm{x}\text{ }=\text{ }\mathrm{\pi }/4\text{ }\mathrm{t}\text{​ }=\text{ }0\\ \mathrm{x}\text{\hspace{0.17em}}=\text{ }0\text{ }\mathrm{t}\text{ }=\text{ }-1\end{array}\right]\\ \therefore \text{ }\mathrm{I}=\underset{-1}{\overset{0}{\int }}\frac{1}{9+16\left[1-{\left(\mathrm{t}\right)}^{2}\right]}\text{ }\mathrm{dt}\\ \text{ }=\underset{-1}{\overset{0}{\int }}\frac{1}{25-16{\left(\mathrm{t}\right)}^{2}}\text{ }\mathrm{dt}\\ \text{ }=\underset{-1}{\overset{0}{\int }}\frac{1}{{5}^{2}-{\left(4\mathrm{t}\right)}^{2}}\text{ }\mathrm{dt}\\ \text{ }=\frac{1}{40}{\left[\mathrm{log}\text{ }\left|\frac{5+4\mathrm{t}}{5-4\mathrm{t}}\right|\right]}_{-1}^{0}\\ \text{ }=\frac{1}{40}\left[\mathrm{log}\text{ }\left|1/\frac{1}{9}\right|\right]\\ \text{ }=\frac{1}{40}\mathrm{log}\text{ }\left|9\right|\text{ }\end{array}$

Q.9

$\mathrm{Evaluate}\text{ }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\mathrm{xtanx}}{\mathrm{secx}\text{ }+\text{\hspace{0.17em}}\mathrm{tanx}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{Let}\text{ }\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\mathrm{xtanx}}{\mathrm{secx}\text{ }+\text{\hspace{0.17em}}\mathrm{tanx}}\text{ }\mathrm{dx}\text{ }.\dots \left(1\right)\\ \text{ \hspace{0.17em}}=\text{ }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\left(\mathrm{\pi }-\mathrm{x}\right)\mathrm{tan}\left(\mathrm{\pi }-\mathrm{x}\right)}{\mathrm{sec}\left(\mathrm{\pi }-\mathrm{x}\right)\text{ }+\text{\hspace{0.17em}}\mathrm{tan}\left(\mathrm{\pi }-\mathrm{x}\right)}\text{ }\mathrm{dx}\\ \text{ \hspace{0.17em}}=\text{ }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{-\left(\mathrm{\pi }-\mathrm{x}\right)\mathrm{tanx}}{-\mathrm{secx}-\text{\hspace{0.17em}}\mathrm{tanx}}\text{ }\mathrm{dx}\\ \text{ }\mathrm{I}\text{\hspace{0.17em}}=\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\left(\mathrm{\pi }-\mathrm{x}\right)\mathrm{tanx}}{\mathrm{secx}\text{ }+\text{\hspace{0.17em}}\mathrm{tanx}}\text{ }\mathrm{dx}\text{ }.\dots \left(2\right)\\ \text{ }2\mathrm{I}\text{\hspace{0.17em}}=\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\mathrm{tanx}}{\mathrm{secx}\text{ }+\text{\hspace{0.17em}}\mathrm{tanx}}\text{ }\mathrm{dx}\text{ }\left[\mathrm{adding}\text{ }\mathrm{equation}\text{ }\left(1\right)\text{ }\mathrm{and}\left(2\right)\right]\text{ }\\ \text{ }=\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\mathrm{sinx}}{1\text{ }+\text{\hspace{0.17em}}\mathrm{sinx}}\text{ }\mathrm{dx}\text{ }\\ \text{ }=\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{1+\mathrm{sinx}}{1\text{ }+\text{\hspace{0.17em}}\mathrm{sinx}}\text{ }\mathrm{dx}-\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{1}{1\text{ }+\text{\hspace{0.17em}}\mathrm{sinx}}\text{ }\mathrm{dx}\\ \text{ }=\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\mathrm{dx}-\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{1}{1\text{ }+\text{\hspace{0.17em}}\mathrm{sinx}}×\frac{1\text{ }-\text{\hspace{0.17em}}\mathrm{sinx}}{1\text{ }-\text{\hspace{0.17em}}\mathrm{sinx}}\text{ }\mathrm{dx}\\ \text{ }=\text{ }\mathrm{\pi }{\left[\mathrm{x}\right]}_{0}^{\mathrm{\pi }}-\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{1\text{ }-\text{\hspace{0.17em}}\mathrm{sinx}}{{\mathrm{cos}}^{2}\mathrm{x}}\mathrm{dx}\\ \text{ }=\text{ }{\mathrm{\pi }}^{2}-\text{ }\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}{\mathrm{sec}}^{2}\mathrm{x}\text{ }-\mathrm{secx}\text{ }\mathrm{tanx}\text{ }\mathrm{dx}\\ \text{ }=\text{ }{\mathrm{\pi }}^{2}-\text{ }\mathrm{\pi }{\left[\mathrm{tanx}-\mathrm{secx}\right]}_{0}^{\mathrm{\pi }}\\ \text{ }=\text{ }{\mathrm{\pi }}^{2}-\text{ }\mathrm{\pi }\left[\mathrm{tan\pi }-\mathrm{sec\pi }\right]\text{ }+\text{\hspace{0.17em}}\mathrm{\pi }\left[\mathrm{tan}0-\mathrm{sec}0\right]\\ \text{ }=\text{ }{\mathrm{\pi }}^{2}-\text{ }\mathrm{\pi }\left[0-\left(-1\right)\right]\text{ }+\text{\hspace{0.17em}}\mathrm{\pi }\left[0-1\right]\\ \text{ }2\mathrm{I}=\text{ }{\mathrm{\pi }}^{2}-\text{ }2\mathrm{\pi }\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\text{ }\frac{{\mathrm{\pi }}^{2}-\text{ }2\mathrm{\pi }}{2}\text{ }=\text{ }\frac{\mathrm{\pi }}{2}\text{ }\left(\mathrm{\pi }-2\right)\end{array}$

Q.10

$\mathrm{Evaluate}\text{ }\underset{1}{\overset{4}{\int }}\left(\left|\mathrm{x}-1\right|+\left|\mathrm{x}-2\right|+\left|\mathrm{x}-3\right|\right)\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{I}\text{ }=\text{ }\underset{1}{\overset{4}{\int }}\left(\left|\mathrm{x}-1\right|+\left|\mathrm{x}-2\right|+\left|\mathrm{x}-3\right|\right)\\ \mathrm{Since}\text{ }\mathrm{f}\left(\mathrm{x}\right)\text{ }=\text{ }\left\{\begin{array}{l}-\left(\mathrm{x}-1\right)-\left(\mathrm{x}-2\right)-\left(\mathrm{x}-3\right)\text{ }\mathrm{x}<1\\ \left(\mathrm{x}-1\right)-\left(\mathrm{x}-2\right)-\left(\mathrm{x}-3\right)\text{ }1\text{ }\le \mathrm{x}<2\\ \left(\mathrm{x}-1\right)+\left(\mathrm{x}-2\right)-\left(\mathrm{x}-3\right)\text{ }2\text{ }\le \mathrm{x}<3\\ \left(\mathrm{x}-1\right)+\left(\mathrm{x}-2\right)+\left(\mathrm{x}-3\right)\text{ }3\le \mathrm{x}\end{array}\right\}\\ \text{ }=\text{ }\left\{\begin{array}{l}-3\mathrm{x}+6;\text{ }\mathrm{x}<1\\ -\mathrm{x}+4;\text{ }1\text{ }\le \mathrm{x}<2\\ \mathrm{x};\text{ }2\text{ }\le \mathrm{x}<3\\ 3\mathrm{x}-6\text{ }3\le \mathrm{x}\end{array}\right\}\\ \mathrm{I}\text{ }=\text{ }\underset{1}{\overset{2}{\int }}\left(-\mathrm{x}+4\right)\text{ }\mathrm{dx}\text{ }+\text{ }\underset{2}{\overset{3}{\int }}\left(\mathrm{x}\right)\text{ }\mathrm{dx}\text{ }+\text{ }\underset{3}{\overset{4}{\int }}\left(3\mathrm{x}-6\right)\text{ }\mathrm{dx}\\ \mathrm{I}\text{ }=\text{ }{\left[-\frac{{\mathrm{x}}^{2}}{2}\text{ }+\text{ }4\mathrm{x}\right]}_{1}^{2}\text{ }+\text{ }{\left[\frac{{\mathrm{x}}^{2}}{2}\right]}_{2}^{3}\text{ }+\text{ }{\left[3\frac{{\mathrm{x}}^{2}}{2}\text{ }-\text{ }6\mathrm{x}\right]}_{3}^{4}\text{ }\\ \mathrm{I}\text{ }=\text{ }\left[\frac{-{2}^{2}}{2}\text{​ }+\text{ }4\left(2\right)\text{ }-\text{ }\frac{-{\left(1\right)}^{2}}{2}-4\left(1\right)\right]\text{ }+\left[\frac{{3}^{2}}{2}\text{​ }-\text{ }\frac{{2}^{2}}{2}\right]\text{ }+\text{ }\left[3\frac{{4}^{2}}{2}\text{​ }-\text{ }6\left(4\right)\text{ }-\text{ }3\frac{{3}^{2}}{2}+6\left(3\right)\right]\\ \mathrm{I}\text{ }=\text{ }-2+8+\frac{1}{2}-4+\frac{9}{2}-2+24-24-\frac{27}{2}+18\\ \text{ }=\text{ }\frac{27}{2}-4\text{ }=\text{ }\frac{27-8}{2}\\ \text{ }=\text{ }\frac{19}{2}\end{array}$

Q.11

$\mathrm{Evaluate}\text{ }\int \frac{\sqrt{\mathrm{tanx}}}{\mathrm{sinx}\text{ }\mathrm{cosx}}\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{\sqrt{\mathrm{tanx}}}{\mathrm{sinx}\text{ }\mathrm{cosx}}\mathrm{dx}\\ =\int \frac{\sqrt{\mathrm{tanx}}}{\frac{\mathrm{sinx}}{\mathrm{cosx}}\text{ }{\mathrm{cos}}^{2}\mathrm{x}}\mathrm{dx}\\ =\int \frac{\sqrt{\mathrm{tanx}}{\mathrm{sec}}^{2}\mathrm{x}}{\mathrm{tanx}}\mathrm{dx}\\ =\int \frac{{\mathrm{sec}}^{2}\mathrm{x}}{\sqrt{\mathrm{tanx}}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{\mathrm{t}}}\text{ }\mathrm{dt}\text{ }=2\sqrt{\mathrm{t}}\text{ }+\mathrm{C}\text{ }\left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{tanx}\text{ }=\text{\hspace{0.17em}}\mathrm{t}\\ {\mathrm{sec}}^{2}\text{ }\mathrm{xdx}\text{ }=\text{\hspace{0.17em}}\mathrm{dt}\end{array}\right]\\ =2\text{ }\sqrt{\mathrm{tanx}}\text{ }+\text{ }\mathrm{C}\end{array}$

Q.12

$\mathrm{Evaluate}\text{ }\int \mathrm{tan}2\mathrm{x}\mathrm{dx}.$

Ans

$\int \mathrm{tan}2\mathrm{xdx}\text{ }=\text{ }\frac{1}{2}\mathrm{log}\left|\mathrm{sec}2\mathrm{x}\right|\text{ }+\text{ }\mathrm{C}$

Q.13

$\mathrm{Evaluate}\text{ }\int \frac{{\mathrm{x}}^{3}\text{ }-\text{ }{\mathrm{x}}^{2}\text{ }+\text{ }\mathrm{x}\text{ }-1}{\mathrm{x}-1}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{{\mathrm{x}}^{3}\text{ }-\text{ }{\mathrm{x}}^{2}\text{ }+\text{ }\mathrm{x}\text{ }-1}{\mathrm{x}-1}\text{ }\mathrm{dx}\\ =\int \frac{{\mathrm{x}}^{2}\left(\mathrm{x}-1\right)\text{ }+\text{ }\left(\mathrm{x}\text{ }-1\right)}{\mathrm{x}-1}\text{ }\mathrm{dx}\\ =\int \frac{\left({\mathrm{x}}^{2}+1\right)\text{ }+\text{ }\left(\mathrm{x}\text{ }-1\right)}{\mathrm{x}-1}\text{ }\mathrm{dx}\\ =\int {\mathrm{x}}^{2}+1\text{ }\mathrm{dx}\text{ }=\text{ }\frac{{\mathrm{x}}^{3}}{3}\text{ }+\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{C}\end{array}$

Q.14

$\mathrm{Evaluate}\text{ }\int \frac{{\mathrm{x}}^{2}}{{\mathrm{x}}^{4}+1}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{{\mathrm{x}}^{2}}{{\mathrm{x}}^{4}+1}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{2{\mathrm{x}}^{2}}{{\mathrm{x}}^{4}+1}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{{\mathrm{x}}^{2}+1+{\mathrm{x}}^{2}-1}{{\mathrm{x}}^{4}+1}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{{\mathrm{x}}^{2}+1}{{\mathrm{x}}^{4}+1}\text{ }\mathrm{dx}\text{ }+\text{ }\frac{1}{2}\int \frac{{\mathrm{x}}^{2}-1}{{\mathrm{x}}^{4}+1}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{1+\frac{1}{{\mathrm{x}}^{2}}}{{\mathrm{x}}^{2}+\frac{1}{{\mathrm{x}}^{2}}}\text{ }\mathrm{dx}\text{ }+\text{ }\frac{1}{2}\int \frac{1-\frac{1}{{\mathrm{x}}^{2}}}{{\mathrm{x}}^{2}+\frac{1}{{\mathrm{x}}^{2}}}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{1+\frac{1}{{\mathrm{x}}^{2}}}{\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+2}\text{ }\mathrm{dx}\text{ }+\text{ }\frac{1}{2}\int \frac{1-\frac{1}{{\mathrm{x}}^{2}}}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)-2}\text{ }\mathrm{dx}\\ \left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{x}-\frac{1}{\mathrm{x}}\text{ }=\text{ }\mathrm{u}\text{ \hspace{0.17em} }\mathrm{x}+\frac{1}{\mathrm{x}}=\text{ }\mathrm{v}\\ \left(\mathrm{x}+\frac{1}{{\mathrm{x}}^{2}}\right)\mathrm{dx}\text{ }=\text{ }\mathrm{du}\text{ }\left(\mathrm{x}-\frac{1}{{\mathrm{x}}^{2}}\right)\mathrm{dx}\text{ }=\text{ }\mathrm{du}\end{array}\right]\\ \therefore \text{ }\int \frac{{\mathrm{x}}^{2}}{{\mathrm{x}}^{4}+1}\text{ }\mathrm{dx}\text{ }=\text{ }\frac{1}{2}\int \frac{1}{{\left(\mathrm{u}\right)}^{2}\text{ }+\text{ }{\left(\sqrt{2}\right)}^{2}}\mathrm{du}\text{ }+\frac{1}{2}\int \frac{1}{{\left(\mathrm{v}\right)}^{2}\text{ }+\text{ }{\left(\sqrt{2}\right)}^{2}}\mathrm{du}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{ }\frac{1}{2}\text{ }\frac{1}{\sqrt{2}}\text{ }{\mathrm{tan}}^{-1}\text{ }\frac{\mathrm{u}}{\sqrt{2}}\text{ }+\text{ }\frac{1}{2}\text{ }\frac{1}{2\sqrt{2}}\text{ }\mathrm{log}\text{ }\left|\frac{\mathrm{v}-\sqrt{2}}{\mathrm{v}+\sqrt{2}}\right|\text{ }+\mathrm{C}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{ }\frac{1}{2\sqrt{2}}\text{ }{\mathrm{tan}}^{-1}\text{ }\frac{\mathrm{x}\text{ }-\text{ }\frac{1}{\mathrm{x}}}{\sqrt{2}}\text{ }+\text{ }\frac{1}{4\sqrt{2}}\text{ }\mathrm{log}\text{ }\left|\frac{\mathrm{x}\text{ }+\text{ }\frac{1}{\mathrm{x}}-\sqrt{2}}{\mathrm{x}\text{ }+\text{ }\frac{1}{\mathrm{x}}+\sqrt{2}}\right|\text{ }+\mathrm{C}\end{array}$

Q.15

$\mathrm{Evaluate}\text{ }\int \frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{\mathrm{sin}\left[\left(\mathrm{x}+\mathrm{a}\right)-2\mathrm{a}\right]}{\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)\text{ }\mathrm{cos}2\mathrm{a}-\text{ }\mathrm{cos}\text{ }\left(\mathrm{x}+\mathrm{a}\right)\text{ }\mathrm{sin}2\mathrm{a}}{\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \mathrm{cos}2\mathrm{adx}-\text{ }\mathrm{sin}2\mathrm{a}\text{ }\int \mathrm{cot}\text{ }\left(\mathrm{x}+\mathrm{a}\right)\text{ }\mathrm{dx}\\ =\text{ }\mathrm{x}\text{ }\mathrm{cos}2\mathrm{a}\text{ }-\text{ }\mathrm{sin}2\mathrm{a}\text{ }\mathrm{log}\text{ }\left|\mathrm{sin}\text{ }\left(\mathrm{x}+\mathrm{a}\right)\right|\text{ }+\mathrm{C}\end{array}$

Q.16

$\mathrm{Evaluate}\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\mathrm{cos}}^{2}\mathrm{x}}{{\mathrm{cos}}^{2}\mathrm{x}\text{ }+\text{ }4{\mathrm{sin}}^{2}\mathrm{x}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\mathrm{cos}}^{2}\mathrm{x}}{{\mathrm{cos}}^{2}\mathrm{x}\text{ }+\text{ }4{\mathrm{sin}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\mathrm{cos}}^{2}\mathrm{x}}{{\mathrm{cos}}^{2}\mathrm{x}\text{ }+\text{ }4\left(1-{\mathrm{cos}}^{2}\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\mathrm{cos}}^{2}\mathrm{x}}{\text{ }4\text{ }-3{\mathrm{cos}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\\ =-\frac{1}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{-3{\mathrm{cos}}^{2}\mathrm{x}}{\text{ }4\text{ }-3{\mathrm{cos}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\\ =-\frac{1}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{4-3{\mathrm{cos}}^{2}\mathrm{x}-4}{\text{ }4\text{ }-3{\mathrm{cos}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\\ =-\frac{1}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{4-3{\mathrm{cos}}^{2}\mathrm{x}}{\text{ }4\text{ }-3{\mathrm{cos}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\text{ }+\text{ }\frac{4}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{\text{ }4\text{ }-3{\mathrm{cos}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\\ =-\frac{1}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{dx}\text{ }+\text{ }\frac{4}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{\text{ }4\text{ }-3\left(\frac{1+\mathrm{cos}2\mathrm{x}}{2}\right)}\text{ }\mathrm{dx}\\ =-\frac{1}{3}{\left[\mathrm{x}\right]}_{0}^{\frac{\mathrm{\pi }}{2}}\text{ }+\text{ }\frac{4}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{\text{ }8\text{ }-3\text{ }-\text{ }3\mathrm{cos}2\mathrm{x}}\text{ }\mathrm{dx}\\ =-\frac{\mathrm{\pi }}{6}\text{ }+\text{ }\frac{8}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{\text{ }5\text{ }-3\text{ }\left(\frac{1-{\mathrm{tan}}^{2}\mathrm{x}}{1+{\mathrm{tan}}^{2}\mathrm{x}}\right)}\text{ }\mathrm{dx}\\ =-\frac{\mathrm{\pi }}{6}\text{ }+\text{ }\frac{8}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\mathrm{sec}}^{2}\mathrm{x}}{\text{ }5\text{ }\left(1+{\mathrm{tan}}^{2}\mathrm{x}\right)-3\text{ }\left(1-{\mathrm{tan}}^{2}\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =-\frac{\mathrm{\pi }}{6}\text{ }+\text{ }\frac{8}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\mathrm{sec}}^{2}\mathrm{x}}{\text{ }2+8{\mathrm{tan}}^{2}\mathrm{x}}\text{ }\mathrm{dx}=\text{ }-\frac{\mathrm{\pi }}{6}\text{ }+\text{ }\frac{4}{3}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\mathrm{sec}}^{2}\mathrm{x}}{\text{ }1+4{\mathrm{tan}}^{2}\mathrm{x}}\text{ }\mathrm{dx}\\ \text{ }\left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{tanx}\text{ }=\mathrm{t}\text{ }\mathrm{x}\text{ }=\text{ }\mathrm{\pi }/2\text{ }\mathrm{t}\text{ }=\text{ }\infty \\ {\mathrm{sec}}^{2}\mathrm{xdx}\text{ }=\text{ }\mathrm{dt}\text{ }\mathrm{x}\text{ }=\text{ }0\text{ }\mathrm{t}\text{ }=\text{ }0\end{array}\right]\\ =\text{ }-\frac{\mathrm{\pi }}{6}\text{ }+\text{ }\frac{4}{3}\underset{0}{\overset{\infty }{\int }}\frac{1}{1+{\left(2\mathrm{t}\right)}^{2}}\text{ }\mathrm{dt}\text{ }=-\frac{\mathrm{\pi }}{6}\text{ }+\text{ }\frac{2}{3}{\left[{\mathrm{tan}}^{-1}\text{ }2\mathrm{t}\right]}_{0}^{\infty }\\ =-\frac{\mathrm{\pi }}{6}\text{ }+\text{ }\frac{2}{3}\left[{\mathrm{tan}}^{-1}\text{ }\infty -\text{ }{\mathrm{tan}}^{-1}\text{ }0\right]\text{ }=\text{​ }-\text{ }\frac{\mathrm{\pi }}{6}\text{ }+\text{ }\frac{\mathrm{\pi }}{3}\text{ }=\text{ }\frac{\mathrm{\pi }}{6}\end{array}$

Q.17

$\text{Evaluate }\text{∫}\frac{{2}^{\mathrm{x}}+{3}^{\mathrm{x}}}{{6}^{\mathrm{x}}}\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{{2}^{\mathrm{x}}+{3}^{\mathrm{x}}}{{6}^{\mathrm{x}}}\text{ }=\text{ }\int \left(\frac{1}{{2}^{\mathrm{x}}}\text{ }+\text{ }\frac{1}{{3}^{\mathrm{x}}}\right)\text{ }\mathrm{dx}\\ \text{ }=\text{ }\frac{{\left(\frac{1}{2}\right)}^{\mathrm{x}}}{\mathrm{log}\left(\frac{1}{2}\right)}\text{ }+\text{​}\frac{{\left(\frac{1}{3}\right)}^{\mathrm{x}}}{\mathrm{log}\left(\frac{1}{3}\right)}\text{ }+\text{ }\mathrm{C}\end{array}$

Q.18

$\mathrm{Evaluate}\text{ }\int \mathrm{tanx}\text{ }\mathrm{tan}\text{ }2\mathrm{x}\text{ }\mathrm{tan}3\mathrm{x}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{tan}3\mathrm{x}\text{ }=\text{ }\mathrm{tan}\left(2\mathrm{x}+\mathrm{x}\right)\\ \text{ }=\text{ }\frac{\mathrm{tan}2\mathrm{x}+\mathrm{tanx}}{1-\mathrm{tan}2\mathrm{xtanx}}\\ \mathrm{tan}3\mathrm{x}\left(1-\mathrm{tan}2\mathrm{xtanx}\right)=\mathrm{tan}2\mathrm{x}+\mathrm{tanx}\\ \mathrm{tan}3\mathrm{x}-\mathrm{tan}2\mathrm{x}\text{ }-\mathrm{tanx}\text{ }=\text{ }\mathrm{tanxtan}2\mathrm{x}\text{ }\mathrm{tan}3\mathrm{x}\\ \int \mathrm{tanxtan}2\mathrm{xtan}3\mathrm{xdx}\\ =\text{ }\int \mathrm{tan}3\mathrm{x}-\mathrm{tan}2\mathrm{x}-\mathrm{tanxdx}\\ =\text{ }\frac{1}{3}\text{ }\mathrm{log}\text{ }\left|\mathrm{sec}3\mathrm{x}\right|\text{ }-\text{ }\frac{1}{2}\text{ }\mathrm{log}\text{ }\left|\mathrm{sec}2\mathrm{x}\right|\text{ }-\text{ }\mathrm{log}\text{ }\left|\mathrm{secx}\right|\text{ }+\text{ }\mathrm{C}\end{array}$

Q.19

$\mathrm{Evaluate}\text{ }\int {\mathrm{cos}}^{3}\left(\mathrm{ax}\text{ }+\text{ }\mathrm{b}\right)\text{ }\mathrm{sin}\text{ }\left(\mathrm{ax}\text{ }+\text{ }\mathrm{b}\right)\text{ }\mathrm{dx}.\text{ }$

Ans

$\begin{array}{l}\int {\mathrm{cos}}^{3}\left(\mathrm{ax}\text{ }+\text{ }\mathrm{b}\right)\text{ }\mathrm{sin}\text{ }\left(\mathrm{ax}\text{ }+\text{ }\mathrm{b}\right)\text{ }\mathrm{dx}\\ \left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{cos}\left(\mathrm{ax}\text{ }+\text{ }\mathrm{b}\right)\text{ }=\text{ }\mathrm{t}\\ -\mathrm{a}\text{ }\mathrm{sin}\text{ }\left(\mathrm{ax}\text{ }+\text{ }\mathrm{b}\right)\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\end{array}\right]\\ =\text{ }-\frac{1}{\mathrm{a}}\text{ }\int {\mathrm{t}}^{3}\mathrm{dt}\\ =\text{ }-\text{ }\frac{{\mathrm{t}}^{4}}{4\mathrm{a}}\text{ }+\text{ }\mathrm{C}\\ =\text{ }-\text{ }\frac{{\mathrm{cos}}^{4}\text{ }\left(\mathrm{ax}\text{ }+\text{ }\mathrm{b}\right)}{4\mathrm{a}}\text{ }+\text{ }\mathrm{C}\end{array}$

Q.20

$\mathrm{Evaluate}\text{ }\int \mathrm{secx}\text{ }\mathrm{log}\left(\mathrm{sec}\text{ }\mathrm{x}+\text{ }\mathrm{tanx}\right)\mathrm{dx}.\text{ }$

Ans

$\begin{array}{l}\int \mathrm{secx}\text{ }\mathrm{log}\left(\mathrm{sec}\text{ }\mathrm{x}+\text{ }\mathrm{tanx}\right)\mathrm{dx}\\ \left[\begin{array}{l}\mathrm{Let}\text{ }\mathrm{log}\left(\mathrm{secx}\text{ }+\text{ }\mathrm{tanx}\right)\text{ }=\text{ }\mathrm{t}\\ \frac{1}{\mathrm{secx}\text{ }+\text{ }\mathrm{tanx}}\left(\mathrm{secx}\text{ }\mathrm{tanx}\text{ }+\text{ }{\mathrm{sec}}^{2}\mathrm{x}\right)\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\\ \mathrm{secx}\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\end{array}\right]\\ =\text{ }\int \mathrm{t}\mathrm{dt}\\ =\text{ }\frac{{\mathrm{t}}^{2}}{2}\text{ }+\text{ }\mathrm{C}\\ =\text{ }\frac{{\left[\mathrm{log}\left(\mathrm{secx}\text{ }+\text{ }\mathrm{tanx}\right)\right]}^{2}}{2}\text{ }+\text{ }\mathrm{C}\end{array}$

Q.21

$\mathrm{Evaluate}\text{ }\int \frac{2\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+2\right)}\text{ }\mathrm{dx}.\text{ }$

Ans

$\begin{array}{l}\int \frac{2\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+2\right)}\text{ }\mathrm{dx}\\ \left[\begin{array}{l}\mathrm{Let}\text{ }{\mathrm{x}}^{2}\text{ }=\text{ }\mathrm{t}\\ 2\mathrm{xdx}\text{ }=\text{ }\mathrm{dt}\end{array}\right]\\ =\text{ }\int \frac{\mathrm{dt}}{\left(\mathrm{t}\text{ }+\text{ }1\right)\left(\mathrm{t}\text{ }+\text{ }2\right)}\\ =\text{ }\int \frac{\mathrm{dt}}{{\mathrm{t}}^{2}\text{ }+\text{ }3\mathrm{t}\text{ }2}\text{ }\mathrm{dt}\\ =\text{ }\int \frac{1}{{\left(\mathrm{t}\text{ }+\text{ }3/2\right)}^{2}-{\left(1/2\right)}^{2}}\text{ }\mathrm{dt}\\ =\text{ }\frac{1}{2\left(1/2\right)}\text{ }\mathrm{log}\text{ }\left|\frac{\mathrm{t}+1}{\mathrm{t}+2}\right|+\text{ }\mathrm{C}\\ =\text{ }\mathrm{log}\text{ }\left|\frac{{\mathrm{x}}^{2}+1}{{\mathrm{x}}^{2}+2}\right|+\text{ }\mathrm{C}\end{array}$

Q.22

$\mathrm{Evaluate}\text{ }\int \frac{3\mathrm{x}\text{ }+\text{ }5}{{\mathrm{x}}^{3}-\text{ }{\mathrm{x}}^{2}+\mathrm{x}\text{ }+\text{ }1}\text{ }\mathrm{dx}.\text{ }$

Ans

$\begin{array}{l}\int \frac{3\mathrm{x}\text{ }+\text{ }5}{{\mathrm{x}}^{3}-\text{ }{\mathrm{x}}^{2}+\mathrm{x}\text{ }+\text{ }1}\text{ }\mathrm{dx}\\ =\int \frac{3\mathrm{x}\text{ }+\text{ }5}{{\mathrm{x}}^{2}\left(\mathrm{x}-1\right)\text{ }-1\left(\mathrm{x}-1\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{3\mathrm{x}\text{ }+\text{ }5}{\left({\mathrm{x}}^{2}-1\right)\left(\mathrm{x}-1\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{3\mathrm{x}\text{ }+\text{ }5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}\text{ }\mathrm{dx}\\ \mathrm{Let}\text{ }\frac{3\mathrm{x}\text{ }+\text{ }5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}\text{ }=\text{ }\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}\text{ }+\text{ }\frac{\mathrm{B}}{{\left(\mathrm{x}-1\right)}^{2}}\text{ }+\text{ }\frac{\mathrm{C}}{\left(\mathrm{x}+1\right)}\\ 3\mathrm{x}\text{ }+\text{ }5\text{​ }=\text{ }\mathrm{A}\left({\mathrm{x}}^{2}-1\right)\text{ }+\text{ }\mathrm{B}\text{ }\left(\mathrm{x}+1\right)\text{ }+\text{ }\mathrm{C}\text{ }{\left(\mathrm{x}-1\right)}^{2}\text{ }\left(\mathrm{i}\right)\\ \mathrm{Put}\text{ }\mathrm{x}\text{ }=\text{ }1\text{ }\mathrm{in}\text{ }\left(\mathrm{i}\right)\\ 8\text{ }=\text{ }2\mathrm{B}\text{ }⇒\text{ }\mathrm{B}\text{ }=\text{ }4\\ \mathrm{Put}\text{ }\mathrm{x}\text{ }=\text{ }-1\text{ }\mathrm{in}\text{ }\left(\mathrm{i}\right)\\ 2\text{ }=\text{ }4\mathrm{C}\text{ }⇒\text{ }\mathrm{C}\text{ }=\text{ }1/2\\ \mathrm{Put}\text{ }\mathrm{x}\text{ }=\text{ }0\text{ }\mathrm{in}\text{ }\left(\mathrm{i}\right)\\ 5\text{ }=\text{ }-\mathrm{A}\text{ }+\text{ }4\text{ }+\text{ }1/2\\ \mathrm{A}\text{ }=\text{​ }-1/2\\ \int \frac{3\mathrm{x}\text{ }+\text{ }5}{{\left(\mathrm{x}\text{ }-\text{ }1\right)}^{2}\left(\mathrm{x}\text{ }+\text{ }1\right)}\text{ }\mathrm{dx}\text{ }=\text{ }\int \frac{-1/2}{\left(\mathrm{x}\text{ }-\text{ }1\right)}\text{ }+\text{​}\frac{4}{{\left(\mathrm{x}\text{ }-\text{ }1\right)}^{2}}\text{ }+\text{ }\frac{1/2}{\left(\mathrm{x}\text{ }+\text{ }1\right)}\text{ }\mathrm{dx}\\ \text{ }=\text{ }-\frac{1}{2}\text{ }\mathrm{log}\text{ }\left|\mathrm{x}-1\right|\text{ }-\text{ }\frac{4}{\mathrm{x}-1}\text{ }+\frac{1}{2}\mathrm{log}\text{ }\left|\mathrm{x}+1\right|+\text{ }\mathrm{C}\\ \text{ }=\text{ }-\text{ }\frac{4}{\mathrm{x}-1}\text{ }+\frac{1}{2}\mathrm{log}\text{ }\left|\frac{\mathrm{x}+1}{\mathrm{x}-1}\right|+\text{ }\mathrm{C}\end{array}$

Q.23

$\mathrm{Evaluate}\text{ }\int \frac{1}{\mathrm{cos}\text{ }\left(\mathrm{x}-\mathrm{a}\right)\text{ }\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\text{ }\mathrm{dx}.\text{ }$

Ans

$\begin{array}{l}\int \frac{1}{\mathrm{cos}\text{ }\left(\mathrm{x}-\mathrm{a}\right)\text{ }\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\int \frac{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}{\mathrm{cos}\text{ }\left(\mathrm{x}-\mathrm{a}\right)\text{ }\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\int \frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{b}-\left(\mathrm{x}-\mathrm{a}\right)\right)}{\mathrm{cos}\text{ }\left(\mathrm{x}-\mathrm{a}\right)\text{ }\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\int \frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)-\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{sin}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{cos}\text{ }\left(\mathrm{x}-\mathrm{a}\right)\text{ }\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\int \frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{cos}\text{ }\left(\mathrm{x}-\mathrm{a}\right)\text{ }\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\text{ }\mathrm{dx}-\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\int \frac{\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{sin}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{cos}\text{ }\left(\mathrm{x}-\mathrm{a}\right)\text{ }\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\int \mathrm{tan}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{dx}\text{ }-\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\int \mathrm{tan}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{dx}\\ =\text{ }\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\mathrm{log}\left|\mathrm{sec}\text{ }\left(\mathrm{x}-\mathrm{b}\right)\right|\text{ }-\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\mathrm{log}\left|\mathrm{sec}\text{ }\left(\mathrm{x}-\mathrm{a}\right)\right|\text{ }+\text{ }\mathrm{C}\\ =\text{ }\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\mathrm{log}\left|\frac{\mathrm{sec}\text{ }\left(\mathrm{x}-\mathrm{b}\right)}{\mathrm{sec}\text{ }\left(\mathrm{x}-\mathrm{a}\right)}\right|\text{ }+\text{ }\mathrm{C}\end{array}$

Q.24

$\mathrm{Evaluate}\text{ }\int \frac{\mathrm{x}+2}{2{\mathrm{x}}^{2}\text{ }+\text{ }6\mathrm{x}\text{ }+\text{ }5}\text{ }\mathrm{dx}.\text{ }$

Ans

$\begin{array}{l}\int \frac{\mathrm{x}+2}{2{\mathrm{x}}^{2}\text{ }+\text{ }6\mathrm{x}\text{ }+\text{ }5}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{4}\int \frac{4\mathrm{x}+8}{2{\mathrm{x}}^{2}\text{ }+\text{ }6\mathrm{x}\text{ }+\text{ }5}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{4}\int \frac{4\mathrm{x}+6+2}{2{\mathrm{x}}^{2}\text{ }+\text{ }6\mathrm{x}\text{ }+\text{ }5}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{4}\int \frac{4\mathrm{x}+6}{2{\mathrm{x}}^{2}\text{ }+\text{ }6\mathrm{x}\text{ }+\text{ }5}\text{ }\mathrm{dx}\text{ }+\text{ }\frac{1}{4}\int \frac{2}{2{\mathrm{x}}^{2}\text{ }+\text{ }6\mathrm{x}\text{ }+\text{ }5}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{4}\int \frac{1}{\mathrm{t}}\mathrm{dt}\text{ }+\text{ }\frac{1}{4}\int \frac{1}{{\mathrm{x}}^{2}\text{ }+\text{ }3\mathrm{x}\text{ }+\text{ }5/2}\text{ }\mathrm{dx}\text{ }\left[\begin{array}{l}\mathrm{Let}\text{ }2{\mathrm{x}}^{2}+6\mathrm{x}+5\text{ }=\text{ }\mathrm{t}\\ \mathrm{so}\text{​ }\mathrm{that}\text{ }\left(4\mathrm{x}+6\right)\text{ }\mathrm{dx}\text{ }=\text{​ }\mathrm{dt}\end{array}\right]\\ =\text{ }\frac{1}{4}\mathrm{log}\text{ }\left|\mathrm{t}\right|\text{ }+\text{ }\frac{1}{4}\int \frac{1}{{\left(\mathrm{x}+3/2\right)}^{2}\text{ }-\text{ }9/4\text{ }+\text{ }10/4}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{4}\mathrm{log}\text{ }\left|\mathrm{t}\right|\text{ }+\text{ }\frac{1}{4}\int \frac{1}{{\left(\mathrm{x}+\frac{3}{2}\right)}^{2}\text{ }-\text{ }\left(\frac{1}{2}\right)}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{4}\mathrm{log}\text{ }\left|\mathrm{t}\right|\text{ }+\text{ }\frac{1}{4}\text{ }\frac{1}{1/2}\text{ }{\mathrm{tan}}^{-1}\text{ }\frac{\mathrm{x}\text{ }+\text{ }\frac{3}{2}}{\frac{1}{2}}\text{ }+\text{ }\mathrm{C}\\ =\text{ }\frac{1}{4}\mathrm{log}\text{ }\left|2{\mathrm{x}}^{2}+6\mathrm{x}+5\right|\text{ }+\text{ }\frac{1}{2}\text{ }{\mathrm{tan}}^{-1}\text{ }\left(2\mathrm{x}+3\right)\text{ }+\text{ }\mathrm{C}\end{array}$

Q.25

$\mathrm{Evaluate}\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{sin}2\mathrm{x}\text{ }{\mathrm{tan}}^{-1}\left(\mathrm{sinx}\right)\text{ }\mathrm{dx}.\text{ }$

Ans

$\begin{array}{l}\mathrm{Let}\text{ }\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{sin}2\mathrm{x}\text{ }{\mathrm{tan}}^{-1}\left(\mathrm{sinx}\right)\text{ }\mathrm{dx}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}2\mathrm{sinx}\text{ }\mathrm{cosx}\text{ }{\mathrm{tan}}^{-1}\left(\mathrm{sinx}\right)\text{ }\mathrm{dx}\\ \mathrm{Also},\text{ }\mathrm{let}\text{ }\mathrm{sinx}\text{ }=\text{ }\mathrm{t}\text{ }⇒\text{ }\mathrm{cosx}\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\\ \mathrm{When}\text{ }\mathrm{x}\text{ }=\text{ }0,\text{ }\mathrm{t}\text{ }=\text{ }0\text{ }\mathrm{and}\text{ }\mathrm{when}\text{ }\mathrm{x}\text{ }=\frac{\mathrm{\pi }}{2},\text{ }\mathrm{t}\text{ }=\text{ }1\\ ⇒\text{ }\mathrm{I}\text{ }=\text{ }2\underset{0}{\overset{1}{\int }}\mathrm{t}\text{ }{\mathrm{tan}}^{-1}\left(\mathrm{t}\right)\text{ }\mathrm{dt}\text{ }.\dots \left(1\right)\\ \mathrm{Consider},\\ \int \mathrm{t}\text{ }{\mathrm{tan}}^{-1}\text{ }\left(\mathrm{t}\right)\text{ }\mathrm{dt}\text{ }=\text{ }{\mathrm{tan}}^{-1}\text{ }\left(\mathrm{t}\right).\int \mathrm{tdt}\text{ }-\text{ }\int \left\{\frac{\mathrm{d}}{\mathrm{dt}}\text{ }\left({\mathrm{tan}}^{-1}\text{ }\left(\mathrm{t}\right)\right)\int \mathrm{tdt}\right\}\text{ }\mathrm{dt}\\ \text{ \hspace{0.17em}}=\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\text{ }\left(\mathrm{t}\right).\frac{{\mathrm{t}}^{2}}{2}-\int \frac{1}{1+{\mathrm{t}}^{2}}\text{ }.\frac{{\mathrm{t}}^{2}}{2}\text{ }\mathrm{dt}\\ \text{ \hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{t}}^{2}.{\mathrm{tan}}^{-1}\text{ }\left(\mathrm{t}\right)}{2}-\frac{1}{2}\int 1\mathrm{dt}\text{​ }+\frac{1}{2}\int \frac{1}{1\text{ }+\text{ }{\mathrm{t}}^{2}}\text{ }\mathrm{dt}\\ \text{ \hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{t}}^{2}.{\mathrm{tan}}^{-1}\text{ }\left(\mathrm{t}\right)}{2}-\frac{1}{2}\mathrm{t}\text{ }+\text{ }\frac{1}{2}\text{ }{\mathrm{tan}}^{-1}\text{​}\left(\mathrm{t}\right)\\ ⇒\text{ }\mathrm{from}\text{ }\left(1\right)\\ \mathrm{I}\text{ }=\text{ }2\underset{0}{\overset{1}{\int }}\mathrm{t}\text{ }{\mathrm{tan}}^{-1}\left(\mathrm{t}\right)\text{ }\mathrm{dt}\text{ }=\text{ }2\left[\frac{{\mathrm{t}}^{2}.{\mathrm{tan}}^{-1}\text{ }\left(\mathrm{t}\right)}{2}-\frac{1}{2}\mathrm{t}\text{ }+\text{ }\frac{1}{2}\text{ }{\mathrm{tan}}^{-1}\text{​}\left(\mathrm{t}\right)\right]\\ =\text{ }\frac{\mathrm{\pi }}{4}\text{ }-1\text{ }+\text{ }\frac{\mathrm{\pi }}{4}\text{ }\\ =\text{ }\frac{\mathrm{\pi }}{2}\text{ }-1.\end{array}$

Q.26

$\mathrm{Evaluate}\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{log}\text{ }\mathrm{tanx}\text{ }\mathrm{dx}.\text{ }$

Ans

$\begin{array}{l}\mathrm{Let}\text{ }\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{log}\text{ }\mathrm{tanx}\text{ }\mathrm{dx}\text{ }.\dots .\left(1\right)\\ ⇒\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{log}\text{ }\mathrm{tan}\left(\frac{\mathrm{\pi }}{2}\text{ }-\mathrm{x}\right)\text{ }\mathrm{dx}\\ ⇒\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{log}\text{ }\mathrm{cotx}\text{ }\mathrm{dx}\text{ }.\dots \left(2\right)\\ \mathrm{On}\text{ }\mathrm{adding}\text{ }\left(1\right)\text{ }\mathrm{and}\text{\hspace{0.17em}}\left(2\right)\\ ⇒2\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left(\mathrm{log}\text{ }\mathrm{tanx}\text{ }+\text{ }\mathrm{log}\text{ }\mathrm{cot}\text{ }\mathrm{x}\right)\text{ }\mathrm{dx}\\ ⇒2\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left(\mathrm{log}\text{ }\mathrm{tanx}.\mathrm{cot}\text{ }\mathrm{x}\right).\text{ }\mathrm{dx}\\ ⇒2\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left(\mathrm{log}1\right).\text{ }\mathrm{dx}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left(0\right).\text{ }\mathrm{dx}\\ ⇒2\mathrm{I}\text{ }=\text{ }0\\ ⇒\mathrm{I}\text{ }=\text{ }0\end{array}$

Q.27

$\mathrm{Evaluate}\text{ }\underset{0}{\overset{1.5}{\int }}\left[{\mathrm{x}}^{2}\right]\text{ }\mathrm{dx},\text{ }\mathrm{where}\text{ }\left[\mathrm{x}\right]\text{ }\mathrm{is}\text{ }\mathrm{greatest}\text{ }\mathrm{integer}\text{ }\mathrm{function}.\text{ }$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\\ \underset{0}{\overset{1.5}{\int }}\left[{\mathrm{x}}^{2}\right]\text{ }\mathrm{dx}\text{ }=\underset{0}{\overset{1}{\int }}\left[{\mathrm{x}}^{2}\right]\text{ }\mathrm{dx}+\underset{1}{\overset{\sqrt{2}}{\int }}\left[{\mathrm{x}}^{2}\right]\text{ }\mathrm{dx}+\underset{\sqrt{2}}{\overset{1.5}{\int }}\left[{\mathrm{x}}^{2}\right]\text{ }\mathrm{dx}\\ \text{ }=\underset{0}{\overset{1}{\int }}0\text{ }\mathrm{dx}+\underset{1}{\overset{\sqrt{2}}{\int }}1\text{ }\mathrm{dx}+\underset{\sqrt{2}}{\overset{1.5}{\int }}2\text{ }\mathrm{dx}\\ \text{ }=0+\left(\sqrt{2}-1\right)+\left(1.5\text{ }-\text{ }\sqrt{2}\right)\\ \text{ }=\sqrt{2}\text{ }-\text{ }1\text{ }+\text{ }3\text{ }-\text{ }2\sqrt{2}\\ \text{ }=2-\sqrt{2}\end{array}$

Q.28

$\mathrm{Evaluate}\text{ }\int {\mathrm{tan}}^{-1}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\\ \mathrm{Evaluate}\text{ }\int {\mathrm{tan}}^{-1}\mathrm{x}.\mathrm{dx}\text{ }\left[\mathrm{Using}\text{ }\mathrm{integration}\text{ }\mathrm{by}\text{ }\mathrm{part}\right]\\ ={\mathrm{tan}}^{-1}\text{​}\mathrm{x}\int \mathrm{d}\mathrm{x}\text{ }-\text{ }\int \left[\int \mathrm{dx}\right]\text{ }\mathrm{d}\text{ }\left({\mathrm{tan}}^{-1}\text{​}\mathrm{x}\right)\\ =\text{ }{\mathrm{xtan}}^{-1}\text{​}\mathrm{x}\text{ }-\int \frac{\mathrm{x}}{1+{\mathrm{x}}^{2}}.\mathrm{dx}\\ \mathrm{Put}\text{ }1+{\mathrm{x}}^{2}\text{ }=\mathrm{t}\\ \mathrm{On}\text{ }\mathrm{differentiating},\\ 2\mathrm{xdx}\text{ }=\text{ }\mathrm{dt}\\ ⇒\text{ }\mathrm{xdx}\text{ }=\text{ }\frac{\mathrm{dt}}{2}\\ =\text{ }{\mathrm{xtan}}^{-1}\text{​}\mathrm{x}\text{ }-\frac{1}{2}\int \frac{\mathrm{dt}}{\mathrm{t}}\\ =\text{ }{\mathrm{xtan}}^{-1}\text{​}\mathrm{x}\text{ }-\frac{1}{2}\mathrm{log}\left|\mathrm{t}\right|\text{ }+\text{ }\mathrm{c}\\ =\text{ }{\mathrm{xtan}}^{-1}\text{​}\mathrm{x}\text{ }-\frac{1}{2}\mathrm{log}\left|1+{\mathrm{x}}^{2}\right|\text{ }+\text{ }\mathrm{c}\end{array}$

Q.29

$\mathrm{Evaluate}\text{ }\int \frac{\mathrm{sinx}}{\sqrt{1+\mathrm{sin}2\mathrm{x}}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{\mathrm{sinx}}{\sqrt{1+\mathrm{sin}2\mathrm{x}}}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{2\mathrm{sinx}}{\sqrt{1+\mathrm{sin}2\mathrm{x}}}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\text{ }+\text{ }\left(\mathrm{sinx}\text{​ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\sqrt{{\left(\mathrm{sinx}\text{​ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}^{2}}}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\text{ }+\text{ }\left(\mathrm{sinx}\text{​ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\text{ }+\text{ }\int \frac{\left(\mathrm{sinx}\text{ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \mathrm{dx}\text{ }+\text{ }\int \frac{\left(\mathrm{sinx}\text{ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ \mathrm{Put}\text{ }\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\text{ }=\text{ }\mathrm{t}\\ \mathrm{On}\text{ }\mathrm{differentiating},\\ =\text{ }\left(\mathrm{sinx}\text{ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\\ =\text{ }\mathrm{x}\text{ }-\text{ }\int \frac{\mathrm{dt}}{\mathrm{t}}\\ =\text{ }\mathrm{x}\text{ }-\text{ }\mathrm{log}\left|\mathrm{t}\right|\text{ }+\text{ }\mathrm{c}\\ =\text{ }\mathrm{x}\text{ }-\text{ }\mathrm{log}\left|\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\right|\text{ }+\text{ }\mathrm{c}\end{array}$

Q.30

$\mathrm{Evaluate}\text{ }\int \frac{1}{1\text{ }+\text{ }\mathrm{cotx}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{1}{1\text{ }+\text{ }\mathrm{cotx}}\text{ }\mathrm{dx}\\ \int \frac{1}{1\text{ }+\text{ }\frac{\mathrm{cosx}}{\mathrm{sinx}}}\text{ }\\ =\text{ }\int \frac{\mathrm{sinx}}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{2\mathrm{sinx}}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\text{ }\frac{1}{2}\int \frac{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\text{ }+\text{ }\left(\mathrm{sinx}\text{​ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\text{ }+\text{ }\int \frac{\left(\mathrm{sinx}\text{ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \mathrm{dx}\text{ }+\text{ }\int \frac{\left(\mathrm{sinx}\text{ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ \mathrm{Put}\text{ }\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\text{ }=\text{ }\mathrm{t}\\ \mathrm{On}\text{ }\mathrm{differentiating},\\ =\text{ }-\left(\mathrm{sinx}\text{ }-\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\\ =\text{ }\mathrm{x}\text{ }-\text{ }\int \frac{\mathrm{dt}}{\mathrm{t}}\\ =\text{ }\mathrm{x}\text{ }-\text{ }\mathrm{log}\left|\mathrm{t}\right|\text{ }+\text{ }\mathrm{c}\\ =\text{ }\mathrm{x}\text{ }-\text{ }\mathrm{log}\left|\left(\mathrm{sinx}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)\right|\text{ }+\text{ }\mathrm{c}\end{array}$

Q.31

$\mathrm{Evaluate}\text{ }\int \frac{{\mathrm{sin}}^{-1}\mathrm{x}}{{\mathrm{sin}}^{-1}\sqrt{\mathrm{x}}+\text{ }{\mathrm{cos}}^{-1}\sqrt{\mathrm{x}}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\\ \int \frac{{\mathrm{sin}}^{-1}\mathrm{x}}{{\mathrm{sin}}^{-1}\sqrt{\mathrm{x}}+\text{ }{\mathrm{cos}}^{-1}\sqrt{\mathrm{x}}}\text{ }\mathrm{dx}\\ \int \frac{{\mathrm{sin}}^{-1}\mathrm{x}}{\frac{\mathrm{\pi }}{2}}\text{ }\mathrm{dx}\left[\because \text{​ }{\mathrm{sin}}^{-1}\text{ }\sqrt{\mathrm{x}}\text{​​​ ​}+\text{ }{\mathrm{cos}}^{-1}\sqrt{\mathrm{x}}\text{ }=\text{ }\frac{\mathrm{\pi }}{2}\right]\\ =\frac{\mathrm{\pi }}{2}\int {\mathrm{sin}}^{-1}\mathrm{x}.\mathrm{dx}\left[\mathrm{Using}\text{ }\mathrm{Integration}\text{ }\mathrm{by}\text{ }\mathrm{part}\right]\\ =\text{\hspace{0.17em} }{\mathrm{sin}}^{-1}\mathrm{x}\int \mathrm{dx}\text{ }-\text{ }\int \left[\int \mathrm{dx}\right]\mathrm{d}\text{ }\left({\mathrm{sin}}^{-1}\mathrm{x}\right)\\ =\text{ }{\mathrm{xsin}}^{-1}\mathrm{x}-\int \frac{\mathrm{xdx}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{ }\\ \mathrm{Put}\text{ }1-{\mathrm{x}}^{2}\text{ }=\text{ }{\mathrm{t}}^{2}\\ \mathrm{On}\text{ }\mathrm{differentiating}\\ -2\mathrm{xdx}\text{ }=\text{ }2\mathrm{tdt}\\ ⇒\text{ }\mathrm{xdx}\text{ }=\text{ }-\mathrm{tdt}\\ =\text{ }{\mathrm{xsin}}^{-1}\mathrm{x}\text{ }+\text{ }\int \frac{\mathrm{tdt}}{\mathrm{t}}\\ =\text{ }{\mathrm{xsin}}^{-1}\mathrm{x}\text{ }+\text{ }\int \mathrm{dt}\\ =\text{ }{\mathrm{xsin}}^{-1}\mathrm{x}\text{ }+\text{ }\mathrm{t}\text{ }+\mathrm{c}\\ =\text{ }{\mathrm{xsin}}^{-1}\mathrm{x}\text{ }+\text{ }\sqrt{1-{\mathrm{x}}^{2}}\text{ }+\text{ }\mathrm{c}\end{array}$

Q.32

$\mathrm{Evaluate}\text{ }\int {\mathrm{e}}^{\mathrm{x}}\left[\sqrt{\mathrm{x}}\text{ }+\text{​ }\frac{1}{2\sqrt{\mathrm{x}}}\right]\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\\ \int {\mathrm{e}}^{\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\text{ }+\text{ }\mathrm{f}‘\left(\mathrm{x}\right)\right]\text{ }=\text{ }{\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\text{ }+\text{ }\mathrm{c}\\ \mathrm{Here},\text{ }\mathrm{f}\left(\mathrm{x}\right)\text{ }=\text{ }\sqrt{\mathrm{x}}\\ ⇒\text{ }\mathrm{f}‘\left(\mathrm{x}\right)\text{ }=\text{ }\frac{1}{2\sqrt{\mathrm{x}}}\\ ⇒\text{ }\int {\mathrm{e}}^{\mathrm{x}}\text{ }\left[\sqrt{\mathrm{x}}\text{ }+\text{ }\frac{1}{2\sqrt{\mathrm{x}}}\right]\mathrm{dx}\text{ }=\text{ }{\mathrm{e}}^{\mathrm{x}}\sqrt{\mathrm{x}}+\mathrm{c}\end{array}$

Q.33

$\mathrm{Evaluate}\text{ }\int \frac{1}{1\text{ }+\text{ }{\mathrm{x}}^{4}}\text{ }\mathrm{d}{\left(\mathrm{x}\right)}^{2}.$

Ans

$\begin{array}{l}\int \frac{\mathrm{d}{\left(\mathrm{x}\right)}^{2}}{1\text{ }+\text{ }{\mathrm{x}}^{4}}\text{ }=\text{ }\int \frac{\mathrm{d}{\left(\mathrm{x}\right)}^{2}}{1\text{ }+\text{ }{\left({\mathrm{x}}^{2}\right)}^{2}}\\ \mathrm{Put}\text{ }{\mathrm{x}}^{2}\text{ }=\text{ }\mathrm{t}\\ \int \frac{\mathrm{dt}}{1\text{ }+\text{ }{\mathrm{t}}^{2}}\text{ }=\text{ }{\mathrm{tan}}^{-1\text{ }}\mathrm{t}\text{ }+\mathrm{c}\\ \text{ }=\text{ }{\mathrm{tan}}^{-1}\text{ }\left({\mathrm{x}}^{2}\right)\text{ }+\mathrm{c}\end{array}$

Q.34

$\mathrm{Evaluate}\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\mathrm{sin}}^{2}\mathrm{x}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{Let}\text{ }\mathrm{I}\text{ }=\text{​}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\mathrm{sin}}^{2}\mathrm{x}\text{ }\mathrm{dx}\text{ }...\left(1\right)\\ ⇒\text{ }\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\mathrm{sin}}^{2}\left(\frac{\mathrm{\pi }}{2}\text{ }-\mathrm{x}\right)\text{ }\mathrm{dx}\\ ⇒\text{ }\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\mathrm{cos}}^{2}\mathrm{x}\text{ }\mathrm{dx}\text{ }.\dots \left(2\right)\\ \mathrm{On}\text{ }\mathrm{adding}\text{ }\left(1\right)\text{ }\mathrm{and}\text{ }\left(2\right),\\ 2\mathrm{I}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left({\mathrm{sin}}^{2}\mathrm{x}\text{ }+\text{ }{\mathrm{cos}}^{2}\mathrm{x}\right)\text{ }\mathrm{dx}\text{ }=\text{ }\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}1.\text{ }\mathrm{dx}\\ ⇒2\mathrm{I}\text{ }=\text{ }{\left[\mathrm{x}\right]}_{0}^{\frac{\mathrm{\pi }}{2}}\text{ }=\text{ }\frac{\mathrm{\pi }}{2}\text{ }-\text{ }0\\ ⇒\mathrm{I}\text{ }=\text{ }\frac{\mathrm{\pi }}{4}\end{array}$

Q.35

$\mathrm{Evaluate}\text{ }\int \frac{1}{{\mathrm{e}}^{\mathrm{x}}\text{ }+\text{ }{\mathrm{e}}^{-\mathrm{x}}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{1}{{\mathrm{e}}^{\mathrm{x}}\text{ }+\text{ }{\mathrm{e}}^{-\mathrm{x}}}\text{ }\mathrm{dx}\\ \mathrm{On}\text{ }\mathrm{multiplying}\text{ }{\mathrm{N}}^{\mathrm{r}}\text{ }\mathrm{and}\text{ }{\mathrm{D}}^{\mathrm{r}}\text{ }\mathrm{by}\text{ }{\mathrm{e}}^{\mathrm{x}},\\ =\int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{2\mathrm{x}}\text{ }+\text{ }1}\text{ }\mathrm{dx}\\ =\int \frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}}{1\text{ }+\text{​}{\left({\mathrm{e}}^{\mathrm{x}}\right)}^{2}}\\ \mathrm{Put}\text{ }{\mathrm{e}}^{\mathrm{x}}\text{ }=\text{ }\mathrm{t}\\ \mathrm{On}\text{ }\mathrm{differentiating},\\ {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\\ =\text{ }\int \frac{\mathrm{dt}}{1\text{ }+\text{​}{\left(\mathrm{t}\right)}^{2}}\\ =\text{ }{\mathrm{tan}}^{-1}\text{​ }\mathrm{t}\text{ }+\mathrm{c}\\ =\text{ }{\mathrm{tan}}^{-1}\text{​ }{\mathrm{e}}^{\mathrm{x}}\text{ }+\mathrm{c}\end{array}$

Q.36

$\mathrm{Prove}\text{ }\mathrm{that}\text{ }\int {\mathrm{e}}^{\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\text{ }+\mathrm{f}‘\left(\mathrm{x}\right)\right]\text{ }=\text{ }{\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\text{ }+\text{ }\mathrm{c}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\\ \int {\mathrm{e}}^{\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\text{ }+\mathrm{f}‘\left(\mathrm{x}\right)\right]\text{ }\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\text{ }\mathrm{f}\left(\mathrm{x}\right)\text{ }\mathrm{dx}\text{ }+\text{ }\int {\mathrm{e}}^{\mathrm{x}}\text{ }\mathrm{f}‘\left(\mathrm{x}\right)\text{ }\mathrm{dx}\\ \left[\mathrm{Using}\text{ }\mathrm{integration}\text{ }\mathrm{by}\text{ }\mathrm{part}\text{ }\mathrm{in}\text{ }\mathrm{first}\text{ }\mathrm{term}\right]\\ =\text{ }\mathrm{f}\left(\mathrm{x}\right)\text{ }\int {\mathrm{e}}^{\mathrm{x}}\text{ }\mathrm{dx}\text{ }-\text{ }\int \left[\int {\mathrm{e}}^{\mathrm{x}}\text{ }\mathrm{dx}\right]\text{ }\mathrm{d}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}\\ =\text{ }\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{\mathrm{x}}\text{ }-\text{ }\int {\mathrm{e}}^{\mathrm{x}}\text{ }\mathrm{f}‘\left(\mathrm{x}\right)\text{ }\mathrm{dx}\text{ }+\text{ }\mathrm{c}\text{ }+\text{ }\int {\mathrm{e}}^{\mathrm{x}}\text{ }\mathrm{f}‘\left(\mathrm{x}\right)\text{ }\mathrm{dx}\\ =\text{ }{\mathrm{e}}^{\mathrm{x}}\text{ }-\text{ }\mathrm{f}\left(\mathrm{x}\right)\text{ }+\text{ }\mathrm{c}\\ \mathrm{Hence}\text{ }\mathrm{proved}\end{array}$

Q.37

$\mathrm{Evaluate}\text{ }\int {\mathrm{sin}}^{-1}\text{ }\mathrm{x}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\\ \int {\mathrm{sin}}^{-1}\text{ }\mathrm{x}.\text{​ }\mathrm{dx}\text{ }\left[\mathrm{Using}\text{ }\mathrm{Integration}\text{ }\mathrm{by}\text{ }\mathrm{part}\right]\\ =\text{ }{\mathrm{sin}}^{-1}\text{ }\mathrm{x}\int \mathrm{dx}\text{ }-\text{ }\int \left[\frac{\mathrm{d}}{\mathrm{dx}}\text{ }\left({\mathrm{sin}}^{-1}\text{ }\mathrm{x}\right)\text{\hspace{0.17em}}\int \mathrm{dx}\right]\text{ }\mathrm{dx}\\ =\text{ }{\mathrm{xsin}}^{-1}\text{ }\mathrm{x}\text{ }-\text{ }\int \frac{\mathrm{xdx}}{\sqrt{1+{2}^{2}}}\\ \mathrm{Put}\text{ }1-{\mathrm{x}}^{2}\text{ }=\text{ }{\mathrm{t}}^{2}\\ \mathrm{On}\text{ }\mathrm{differentiating}\\ -2\mathrm{xdx}\text{ }=\text{ }2\mathrm{tdt}\\ ⇒\text{ }\mathrm{xdx}\text{ }=\text{ }-\mathrm{tdt}\\ =\text{ }{\mathrm{xsin}}^{-1}\text{ }\mathrm{x}\text{ }+\text{ }\int \frac{\mathrm{tdt}}{\mathrm{t}}\\ =\text{ }{\mathrm{xsin}}^{-1}\text{ }\mathrm{x}\text{ }+\text{ }\int \mathrm{dt}\\ =\text{ }{\mathrm{xsin}}^{-1}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{t}\text{ }+\mathrm{c}\\ =\text{ }{\mathrm{xsin}}^{-1}\text{ }\mathrm{x}\text{ }+\text{ }\sqrt{1-{\mathrm{x}}^{2}}\text{ }+\mathrm{c}\end{array}$

Q.38 Find the anti-derivative of tanx.

Ans

$\begin{array}{l}\mathrm{The}\text{ }\mathrm{anti}-\mathrm{derivative}\text{ }\mathrm{of}\text{​ }\mathrm{tanx}\text{ }=\text{ }\int \mathrm{tan}\text{ }\mathrm{x}\text{ }\mathrm{dx}\\ \text{ }=\text{ }\int \frac{\mathrm{sin}\text{ }\mathrm{x}}{\mathrm{cosx}}\text{ }\mathrm{dx}\\ \mathrm{Put}\text{ }\mathrm{cosx}\text{ }=\mathrm{t}\\ \mathrm{On}\text{ }\mathrm{differentiating}\\ ⇒\text{ }\mathrm{sinx}\text{ }\mathrm{dx}\text{ }=-\mathrm{dt}\\ =-\int \frac{\mathrm{dt}}{\mathrm{t}}\text{ }\\ =\text{ }-\mathrm{log}\left|\mathrm{t}\right|\text{ }+\mathrm{c}\\ =\text{ }-\text{ }\mathrm{log}\text{ }\left|\mathrm{cosx}\right|\text{ }+\text{ }\mathrm{c}\end{array}$

Q.39

$\mathrm{Evaluate}\text{ }\int \mathrm{d}\text{ }\left(\mathrm{sinx}\right).$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\\ \int \mathrm{d}\text{ }\left(\mathrm{sinx}\right)\\ \mathrm{Put}\text{ }\mathrm{sinx}\text{ }=\mathrm{t}\\ \int \mathrm{dt}\text{ }=\text{ }\mathrm{t}\text{​ }+\text{ }\mathrm{c}\\ \text{ }=\text{ }\mathrm{sinx}\text{ }+\mathrm{c}\end{array}$

Q.40

$\mathrm{Evaluate}\text{ }\int \mathrm{log}\text{ }\mathrm{x}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\\ \int \mathrm{log}\text{\hspace{0.17em}}\mathrm{x}.\mathrm{dx}\left[\mathrm{Using}\text{ }\mathrm{Integration}\text{ }\mathrm{by}\text{ }\mathrm{part}\right]\\ =\text{ }\mathrm{log}\text{ }\mathrm{x}.\int \mathrm{d}\mathrm{x}\text{ }-\text{ }\int \left[\int \mathrm{dx}\right]\text{\hspace{0.17em}}\mathrm{d}\text{ }\left(\mathrm{log}\text{ }\mathrm{x}\right)\\ =\text{ }\mathrm{xlog}\text{ }\mathrm{x}\text{ }-\int \overline{)\mathrm{x}}.\frac{1}{\overline{)\mathrm{x}}}.\mathrm{dx}\\ =\text{ }\mathrm{xlog}\text{ }\mathrm{x}\text{ }-\int \mathrm{dx}\\ =\text{ }\mathrm{xlog}\text{ }\mathrm{x}\text{ }-\text{ }\mathrm{x}+\mathrm{c}\end{array}$

Q.41

$\mathrm{Evaluate}\text{ }\underset{-1}{\overset{1}{\int }}\mathrm{logx}\text{ }\mathrm{dx}.$

Ans

We have, 0∈ [-1,1]
Since logx is not defined in [-1,1],
log0 doesn’t exist.
Therefore, logx is non-integrable with limit -1 to 1.

Q.42

$\mathrm{Evaluate}\text{ }\int {\mathrm{sin}}^{2}\mathrm{x}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\int {\mathrm{sin}}^{2}\mathrm{x}\text{ }\mathrm{dx}=\text{ }\frac{1}{2}\int \left(1\text{ }-\text{ }\mathrm{cos}2\mathrm{x}\right)\text{ }\mathrm{dx}\\ \text{ }=\text{ }\frac{1}{2}\left[\int \mathrm{dx}\text{ }-\text{ }\int \mathrm{cos}\text{ }2\mathrm{x}\text{ }\mathrm{dx}\right]\\ \text{ }=\text{ }\frac{1}{2}\left[\mathrm{x}\text{ }-\text{ }\frac{\mathrm{sin}\text{ }2\mathrm{x}}{2}\right]\text{ }+\text{ }\mathrm{c}\\ \text{ }=\text{ }\frac{\mathrm{x}}{2}-\text{ }\frac{\mathrm{sin}\text{ }2\mathrm{x}}{4}\text{ }+\text{ }\mathrm{c}\end{array}$

Q.43

$\mathrm{Evaluate}\text{ }\int \frac{{\mathrm{e}}^{\mathrm{x}-1}\text{ }+\text{ }{\mathrm{x}}^{\mathrm{e}-1}}{{\mathrm{e}}^{\mathrm{x}}\text{ }+\text{ }{\mathrm{x}}^{\mathrm{e}}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\text{ }\int \frac{{\mathrm{e}}^{\mathrm{x}-1}\text{ }+\text{ }{\mathrm{x}}^{\mathrm{e}-1}}{{\mathrm{e}}^{\mathrm{x}}\text{ }+\text{ }{\mathrm{x}}^{\mathrm{e}}}\text{ }\mathrm{dx}.\\ \mathrm{Put}\text{ }{\mathrm{e}}^{\mathrm{x}}\text{ }+\text{ }{\mathrm{x}}^{\mathrm{e}}\text{ }=\text{ }\mathrm{t}\\ \mathrm{On}\text{ }\mathrm{differentiating},\\ \left({\mathrm{e}}^{\mathrm{x}}\text{ }+\text{ }{\mathrm{ex}}^{\mathrm{e}-1}\right)\text{ }\mathrm{dx}=\text{ }\mathrm{dt}\\ ⇒\text{ }\mathrm{e}\left({\mathrm{e}}^{\mathrm{x}-1}\text{ }+\text{ }{\mathrm{x}}^{\mathrm{e}-1}\right)\mathrm{dx}=\text{ }\mathrm{dt}\\ ⇒\text{ }\left({\mathrm{e}}^{\mathrm{x}-1}\text{ }+\text{ }{\mathrm{x}}^{\mathrm{e}-1}\right)\mathrm{dx}=\text{ }\frac{\mathrm{dt}}{\mathrm{e}}\\ \mathrm{Now},\text{ }\frac{1}{\mathrm{e}}\text{ }\int \frac{\mathrm{dt}}{\mathrm{t}}\text{ }=\text{ }\frac{1}{\mathrm{e}}\text{ }\mathrm{log}\text{ }\left|\mathrm{t}\right|\text{ }+\text{ }\mathrm{c}\\ \text{ \hspace{0.17em}}=\text{ }\frac{1}{\mathrm{e}}\text{ }\mathrm{log}\text{ }\left|{\mathrm{e}}^{\mathrm{x}}\text{ }+\text{ }{\mathrm{x}}^{\mathrm{e}}\right|\text{ }+\text{ }\mathrm{c}\end{array}$

Q.44

$\mathrm{Evaluate}\underset{0}{\overset{1.5}{\int }}\left[\mathrm{x}\right]\text{ }\mathrm{dx},\text{ }\mathrm{where}\text{ }\left[\mathrm{x}\right]\text{ }\mathrm{is}\text{ }\mathrm{greatest}\text{ }\mathrm{interger}\text{ }\mathrm{function}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ }\mathrm{have},\text{ }\underset{0}{\overset{1.5}{\int }}\left[\mathrm{x}\right]\text{ }\mathrm{dx}\text{ }=\text{ }\underset{0}{\overset{1}{\int }}\left[\mathrm{x}\right]\text{ }\mathrm{dx}\text{ }+\text{ }\underset{0}{\overset{1.5}{\int }}\left[\mathrm{x}\right]\text{ }\mathrm{dx}\\ \text{ \hspace{0.17em}}=\text{ }\underset{0}{\overset{1}{\int }}0\text{ }\mathrm{dx}\text{ }+\text{ }\underset{0}{\overset{1.5}{\int }}1\text{ }\mathrm{dx}\\ \text{ \hspace{0.17em}}=\text{ }0\text{ }+\text{ }{\left(\mathrm{x}\right)}_{1}^{1.5}\\ \text{ \hspace{0.17em}}=\text{ }1.5\text{ }-\text{ }1\\ \text{ \hspace{0.17em}}=\text{ }0.5\end{array}$

Q.45

$\mathrm{Evaluate}\int \frac{{\mathrm{x}}^{2}}{1\text{ }+\text{ }{\mathrm{x}}^{3}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{{\mathrm{x}}^{2}}{1\mathrm{ }+\mathrm{ }{\mathrm{x}}^{3}}\mathrm{ }\mathrm{dx}\\ \mathrm{Put}\mathrm{ }1\mathrm{ }+\mathrm{ }{\mathrm{x}}^{3}\mathrm{ }=\mathrm{ }\mathrm{t}\\ \mathrm{On}\mathrm{ }\mathrm{differentiating},\\ 3{\mathrm{x}}^{2}\mathrm{ }\mathrm{dx}\mathrm{ }=\mathrm{ }\mathrm{dt}\\ =\mathrm{ }\frac{1}{3}\int \frac{\mathrm{dt}}{\mathrm{t}}\\ =\mathrm{ }\frac{1}{3}\mathrm{log}\mathrm{ }\mathrm{t}\mathrm{ }+\mathrm{ }\mathrm{c}\\ =\mathrm{ }\frac{1}{3}\mathrm{log}\mathrm{ }\left|1\mathrm{ }+\mathrm{ }{\mathrm{x}}^{3}\right|\mathrm{ }+\mathrm{ }\mathrm{c}\end{array}$

Q.46

$\mathrm{Evaluate}\int \frac{\left(\mathrm{sin}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\sqrt{1\text{ }+\text{ }\mathrm{sin}\text{ }2\mathrm{x}}}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \frac{\left(\mathrm{sin}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\sqrt{1\text{ }+\text{ }\mathrm{sin}\text{ }2\mathrm{x}}}\text{ }\mathrm{dx}\\ =\int \frac{\left(\mathrm{sin}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\sqrt{{\left(\mathrm{sin}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}^{2}}}\text{ }\mathrm{dx}\\ =\int \frac{\left(\mathrm{sin}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}{\left(\mathrm{sin}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{cos}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\int \mathrm{dx}\\ =\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{c}\end{array}$

Q.47

$\mathrm{Evaluate}\text{ }\int \mathrm{sec}\text{ }\mathrm{x}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\int \mathrm{sec}\text{ }\mathrm{x}\text{ }\mathrm{dx}\\ \mathrm{On}\text{ }\mathrm{multiplying}\text{ }{\mathrm{N}}^{\mathrm{r}}\text{ }\mathrm{and}\text{ }{\mathrm{D}}^{\mathrm{r}}\text{​ }\mathrm{by}\text{ }\left(\mathrm{sec}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{tan}\text{ }\mathrm{x}\right)\\ =\text{ }\int \frac{\mathrm{sec}\text{ }\mathrm{x}\left(\mathrm{sec}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{tan}\text{ }\mathrm{x}\right)}{\left(\mathrm{sec}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{tan}\text{ }\mathrm{x}\right)}\text{ }\mathrm{dx}\\ =\text{ }\int \frac{\left({\mathrm{sec}}^{2}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{sec}\text{ }\mathrm{x}.\mathrm{tan}\text{ }\mathrm{x}\right)\text{ }\mathrm{dx}}{\left(\mathrm{tan}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{sec}\text{ }\mathrm{x}\right)}\\ \mathrm{Put}\text{ }\mathrm{tan}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{sec}\text{ }\mathrm{x}\text{ }=\text{ }\mathrm{t}\\ \mathrm{On}\text{ }\mathrm{differentiating},\\ \left({\mathrm{sec}}^{2}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{sec}\text{ }\mathrm{x}.\mathrm{tan}\text{ }\mathrm{x}\right)\text{ }\mathrm{dx}\text{ }=\text{ }\mathrm{dt}\\ =\text{ }\int \frac{\mathrm{dt}}{\mathrm{t}}\text{ }=\text{\hspace{0.17em}}\mathrm{log}\text{ }\left|\mathrm{t}\right|\text{ }+\text{ }\mathrm{c}\\ =\text{ }\mathrm{log}\text{ }\left|\mathrm{sec}\text{ }\mathrm{x}\text{ }+\text{ }\mathrm{tan}\text{ }\mathrm{x}\right|\text{ }+\text{ }\mathrm{c}\end{array}$

Q.48

$\mathrm{Evaluate}\text{ }\underset{-\frac{\mathrm{\pi }}{2}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\mathrm{sin}}^{7}\mathrm{x}\text{ }\mathrm{dx}.$

Ans

$\begin{array}{l}\mathrm{Let}\text{ }\mathrm{f}\left(\mathrm{x}\right)\text{ }=\text{ }{\mathrm{sin}}^{7}\text{ }\mathrm{x}\\ \because \text{ }\mathrm{f}\left(-\mathrm{x}\right)\text{ }=\text{ }{\mathrm{sin}}^{7}\text{ }\left(-\mathrm{x}\right)\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} }=\text{ }-{\mathrm{sin}}^{7}\text{ }\mathrm{x}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} }=\text{ }-\mathrm{f}\text{ }\left(\mathrm{x}\right)\\ \therefore \text{ }\mathrm{f}\left(\mathrm{x}\right)\text{ }={\mathrm{sin}}^{7}\text{ }\mathrm{x}\text{ }\mathrm{is}\text{​ }\mathrm{an}\text{ }\mathrm{odd}\text{ }\mathrm{function}.\\ ⇒\text{ }\underset{-\frac{\mathrm{\pi }}{2}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\mathrm{sin}}^{7\text{ }}\text{x }\mathrm{dx}\text{ }=\text{ }0\end{array}$