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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:

[f(x) + g(x) dx] = f(x) dx + g(x) dx
ddx[F(x)]=f(x)
k.f(x) = k. f(x)dx
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:

xndx = xn+1n+1 + C, n -1
1xdx = log x+C
exdx = ex + C
axdx = axlogea+C
sin x dx =-cos x +C
cos x dx=sin x+C
sec2x dx=tan x+C
sec x. tan x dx=sec x+C
cosec x. cot x dx=-cosec x+C
tan x dx= log|cosx|+ C = log|secx|+ C
cot x dx = log|sinx|+ C
sec x dx = log|secx + tanx|+ C
cosec x dx = log|cosecx − cotx|+ C
11-x2dx = sin-1x +C
11+x2dx = tan-1x +C
1xx2-1dx = sec-1x +C

Geometric Interpretation:

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
1x2+a2dx = log x+x2+a2 + C
1x2-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+ 1x4+ kx2+ 1dx or x2- 1x4+ 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

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Q.1

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

Ans

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

Q.2

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

Ans

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

Q.3

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

Ans

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

Q.4

$Evaluate \int_{0}^{1} \frac{1}{\sqrt{1 + x} - \sqrt{x}} dx$

Ans

$\begin{array}{l} \int_{0}^{1} \frac{1}{\sqrt{1 + x} - \sqrt{x}} dx \\ = \int_{0}^{1} \frac{1}{\sqrt{1 + x} - \sqrt{x}} \times \frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}} dx \\ = \int_{0}^{1} \frac{\sqrt{1 + x} + \sqrt{x}}{1} dx = \int_{0}^{1} \sqrt{1 + x} + \sqrt{x} dx \\ = {[\frac{2 {(1 + x)}^{3 / 2}}{3}]}_{0}^{1} + {[\frac{2 {(x)}^{3 / 2}}{3}]}_{0}^{1} \\ = [\frac{2 {(1 + 1)}^{3 / 2}}{3} - \frac{2 {(1 + 0)}^{3 / 2}}{3}] + [\frac{2 {(1)}^{3 / 2}}{3} - \frac{2 {(0)}^{3 / 2}}{3}] \\ = \frac{2 {(1 + 1)}^{3 / 2}}{3} - \frac{2}{3} + \frac{2}{3} \\ = \frac{2 {(2)}^{3 / 2}}{3} = \frac{4 \sqrt{2}}{3} \end{array}$

Q.5

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

Ans

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

Q.6

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

Ans

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

Q.7

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

Ans

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

Q.8

$Evaluate \int_{0}^{π / 4} \frac{sinx + cosx}{9 + 16 \sin 2 x} dx$

Ans

$\begin{array}{l} I = \int_{0}^{π / 4} \frac{sinx + cosx}{9 + 16 \sin 2 x} dx \\ = \int_{0}^{π / 4} \frac{sinx + cosx}{9 + 16 [1 - {(sinx - cosx)}^{2}]} dx \\ [\begin{array}{l} Let sinx - cosx = t \\ (cosx + sinx) dx = dt \\ x = π / 4 t = 0 \\ x = 0 t = - 1 \end{array}] \\ ∴ I = \int_{- 1}^{0} \frac{1}{9 + 16 [1 - {(t)}^{2}]} dt \\ = \int_{- 1}^{0} \frac{1}{25 - 16 {(t)}^{2}} dt \\ = \int_{- 1}^{0} \frac{1}{5^{2} - {(4 t)}^{2}} dt \\ = \frac{1}{40} {[\log |\frac{5 + 4 t}{5 - 4 t}|]}_{- 1}^{0} \\ = \frac{1}{40} [\log |1 / \frac{1}{9}|] \\ = \frac{1}{40} \log |9| \end{array}$

Q.9

$Evaluate \int_{0}^{π} \frac{xtanx}{secx + tanx} dx .$

Ans

$\begin{array}{l} Let I = \int_{0}^{π} \frac{xtanx}{secx + tanx} dx . \dots (1) \\ = \int_{0}^{π} \frac{(π - x) \tan (π - x)}{\sec (π - x) + \tan (π - x)} dx \\ = \int_{0}^{π} \frac{- (π - x) tanx}{- secx - tanx} dx \\ I = π \int_{0}^{π} \frac{(π - x) tanx}{secx + tanx} dx . \dots (2) \\ 2 I = π \int_{0}^{π} \frac{tanx}{secx + tanx} dx [adding equation (1) and (2)] \\ = π \int_{0}^{π} \frac{sinx}{1 + sinx} dx \\ = π \int_{0}^{π} \frac{1 + sinx}{1 + sinx} dx - π \int_{0}^{π} \frac{1}{1 + sinx} dx \\ = π \int_{0}^{π} dx - π \int_{0}^{π} \frac{1}{1 + sinx} \times \frac{1 - sinx}{1 - sinx} dx \\ = π {[x]}_{0}^{π} - π \int_{0}^{π} \frac{1 - sinx}{\cos^{2} x} dx \\ = π^{2} - π \int_{0}^{π} \sec^{2} x - secx tanx dx \\ = π^{2} - π {[tanx - secx]}_{0}^{π} \\ = π^{2} - π [tanπ - secπ] + π [\tan 0 - \sec 0] \\ = π^{2} - π [0 - (- 1)] + π [0 - 1] \\ 2 I = π^{2} - 2 π \\ I = \frac{π^{2} - 2 π}{2} = \frac{π}{2} (π - 2) \end{array}$

Q.10

$Evaluate \int_{1}^{4} (|x - 1| + |x - 2| + |x - 3|) dx .$

Ans

$\begin{array}{l} I = \int_{1}^{4} (|x - 1| + |x - 2| + |x - 3|) \\ Since f (x) = \{\begin{array}{l} - (x - 1) - (x - 2) - (x - 3) x < 1 \\ (x - 1) - (x - 2) - (x - 3) 1 \leq x < 2 \\ (x - 1) + (x - 2) - (x - 3) 2 \leq x < 3 \\ (x - 1) + (x - 2) + (x - 3) 3 \leq x \end{array}\} \\ = \{\begin{array}{l} - 3 x + 6; x < 1 \\ - x + 4; 1 \leq x < 2 \\ x; 2 \leq x < 3 \\ 3 x - 6 3 \leq x \end{array}\} \\ I = \int_{1}^{2} (- x + 4) dx + \int_{2}^{3} (x) dx + \int_{3}^{4} (3 x - 6) dx \\ I = {[- \frac{x^{2}}{2} + 4 x]}_{1}^{2} + {[\frac{x^{2}}{2}]}_{2}^{3} + {[3 \frac{x^{2}}{2} - 6 x]}_{3}^{4} \\ I = [\frac{- 2^{2}}{2} + 4 (2) - \frac{- {(1)}^{2}}{2} - 4 (1)] + [\frac{3^{2}}{2} - \frac{2^{2}}{2}] + [3 \frac{4^{2}}{2} - 6 (4) - 3 \frac{3^{2}}{2} + 6 (3)] \\ I = - 2 + 8 + \frac{1}{2} - 4 + \frac{9}{2} - 2 + 24 - 24 - \frac{27}{2} + 18 \\ = \frac{27}{2} - 4 = \frac{27 - 8}{2} \\ = \frac{19}{2} \end{array}$

Q.11

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

Ans

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

Q.12

$Evaluate \int \tan 2 x dx .$

Ans

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

Q.13

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

Ans

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

Q.14

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

Ans

$\begin{array}{l} \int \frac{x^{2}}{x^{4} + 1} dx \\ = \frac{1}{2} \int \frac{2 x^{2}}{x^{4} + 1} dx \\ = \frac{1}{2} \int \frac{x^{2} + 1 + x^{2} - 1}{x^{4} + 1} dx \\ = \frac{1}{2} \int \frac{x^{2} + 1}{x^{4} + 1} dx + \frac{1}{2} \int \frac{x^{2} - 1}{x^{4} + 1} dx \\ = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx + \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx \\ = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{(x - \frac{1}{x}) + 2} dx + \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{(x + \frac{1}{x}) - 2} dx \\ [\begin{array}{l} Let x - \frac{1}{x} = u x + \frac{1}{x} = v \\ (x + \frac{1}{x^{2}}) dx = du (x - \frac{1}{x^{2}}) dx = du \end{array}] \\ ∴ \int \frac{x^{2}}{x^{4} + 1} dx = \frac{1}{2} \int \frac{1}{{(u)}^{2} + {(\sqrt{2})}^{2}} du + \frac{1}{2} \int \frac{1}{{(v)}^{2} + {(\sqrt{2})}^{2}} du \\ = \frac{1}{2} \frac{1}{\sqrt{2}} \tan^{- 1} \frac{u}{\sqrt{2}} + \frac{1}{2} \frac{1}{2 \sqrt{2}} \log |\frac{v - \sqrt{2}}{v + \sqrt{2}}| + C \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} \frac{x - \frac{1}{x}}{\sqrt{2}} + \frac{1}{4 \sqrt{2}} \log |\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}| + C \end{array}$

Q.15

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

Ans

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

Q.16

$Evaluate \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\cos^{2} x + 4 \sin^{2} x} dx .$

Ans

$\begin{array}{l} \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\cos^{2} x + 4 \sin^{2} x} dx \\ = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\cos^{2} x + 4 (1 - \cos^{2} x)} dx \\ = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{4 - 3 \cos^{2} x} dx \\ = - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{- 3 \cos^{2} x}{4 - 3 \cos^{2} x} dx \\ = - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{4 - 3 \cos^{2} x - 4}{4 - 3 \cos^{2} x} dx \\ = - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{4 - 3 \cos^{2} x}{4 - 3 \cos^{2} x} dx + \frac{4}{3} \int_{0}^{\frac{π}{2}} \frac{1}{4 - 3 \cos^{2} x} dx \\ = - \frac{1}{3} \int_{0}^{\frac{π}{2}} dx + \frac{4}{3} \int_{0}^{\frac{π}{2}} \frac{1}{4 - 3 (\frac{1 + \cos 2 x}{2})} dx \\ = - \frac{1}{3} {[x]}_{0}^{\frac{π}{2}} + \frac{4}{3} \int_{0}^{\frac{π}{2}} \frac{1}{8 - 3 - 3 \cos 2 x} dx \\ = - \frac{π}{6} + \frac{8}{3} \int_{0}^{\frac{π}{2}} \frac{1}{5 - 3 (\frac{1 - \tan^{2} x}{1 + \tan^{2} x})} dx \\ = - \frac{π}{6} + \frac{8}{3} \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{5 (1 + \tan^{2} x) - 3 (1 - \tan^{2} x)} dx \\ = - \frac{π}{6} + \frac{8}{3} \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{2 + 8 \tan^{2} x} dx = - \frac{π}{6} + \frac{4}{3} \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{1 + 4 \tan^{2} x} dx \\ [\begin{array}{l} Let tanx = t x = π / 2 t = \infty \\ \sec^{2} xdx = dt x = 0 t = 0 \end{array}] \\ = - \frac{π}{6} + \frac{4}{3} \int_{0}^{\infty} \frac{1}{1 + {(2 t)}^{2}} dt = - \frac{π}{6} + \frac{2}{3} {[\tan^{- 1} 2 t]}_{0}^{\infty} \\ = - \frac{π}{6} + \frac{2}{3} [\tan^{- 1} \infty - \tan^{- 1} 0] = - \frac{π}{6} + \frac{π}{3} = \frac{π}{6} \end{array}$

Q.17

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

Ans

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

Q.18

$Evaluate \int tanx \tan 2 x \tan 3 x dx .$

Ans

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

Q.19

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

Ans

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

Q.20

$Evaluate \int secx \log (\sec x + tanx) dx .$

Ans

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

Q.21

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

Ans

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

Q.22

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

Ans

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

Q.23

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

Ans

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

Q.24

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

Ans

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

Q.25

$Evaluate \int_{0}^{\frac{π}{2}} \sin 2 x \tan^{- 1} (sinx) dx .$

Ans

$\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \sin 2 x \tan^{- 1} (sinx) dx = \int_{0}^{\frac{π}{2}} 2 sinx cosx \tan^{- 1} (sinx) dx \\ Also, let sinx = t \Rightarrow cosx dx = dt \\ When x = 0, t = 0 and when x = \frac{π}{2}, t = 1 \\ \Rightarrow I = 2 \int_{0}^{1} t \tan^{- 1} (t) dt . \dots (1) \\ Consider, \\ \int t \tan^{- 1} (t) dt = \tan^{- 1} (t) . \int tdt - \int \{\frac{d}{dt} (\tan^{- 1} (t)) \int tdt\} dt \\ = \tan^{- 1} (t) . \frac{t^{2}}{2} - \int \frac{1}{1 + t^{2}} . \frac{t^{2}}{2} dt \\ = \frac{t^{2} . \tan^{- 1} (t)}{2} - \frac{1}{2} \int 1 dt + \frac{1}{2} \int \frac{1}{1 + t^{2}} dt \\ = \frac{t^{2} . \tan^{- 1} (t)}{2} - \frac{1}{2} t + \frac{1}{2} \tan^{- 1} (t) \\ \Rightarrow from (1) \\ I = 2 \int_{0}^{1} t \tan^{- 1} (t) dt = 2 [\frac{t^{2} . \tan^{- 1} (t)}{2} - \frac{1}{2} t + \frac{1}{2} \tan^{- 1} (t)] \\ = \frac{π}{4} - 1 + \frac{π}{4} \\ = \frac{π}{2} - 1 . \end{array}$

Q.26

$Evaluate \int_{0}^{\frac{π}{2}} \log tanx dx .$

Ans

$\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \log tanx dx . \dots. (1) \\ \Rightarrow I = \int_{0}^{\frac{π}{2}} \log \tan (\frac{π}{2} - x) dx \\ \Rightarrow I = \int_{0}^{\frac{π}{2}} \log cotx dx . \dots (2) \\ On adding (1) and (2) \\ \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} (\log tanx + \log \cot x) dx \\ \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} (\log tanx . \cot x) . dx \\ \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} (\log 1) . dx = \int_{0}^{\frac{π}{2}} (0) . dx \\ \Rightarrow 2 I = 0 \\ \Rightarrow I = 0 \end{array}$

Q.27

$Evaluate \int_{0}^{1 .5} [x^{2}] dx, where [x] is greatest integer function .$

Ans

$\begin{array}{l} We have, \\ \int_{0}^{1 .5} [x^{2}] dx = \int_{0}^{1} [x^{2}] dx + \int_{1}^{\sqrt{2}} [x^{2}] dx + \int_{\sqrt{2}}^{1 .5} [x^{2}] dx \\ = \int_{0}^{1} 0 dx + \int_{1}^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{1 .5} 2 dx \\ = 0 + (\sqrt{2} - 1) + (1 .5 - \sqrt{2}) \\ = \sqrt{2} - 1 + 3 - 2 \sqrt{2} \\ = 2 - \sqrt{2} \end{array}$

Q.28

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

Ans

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

Q.29

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

Ans

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

Q.30

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

Ans

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

Q.31

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

Ans

$\begin{array}{l} We have, \\ \int \frac{\sin^{- 1} x}{\sin^{- 1} \sqrt{x} + \cos^{- 1} \sqrt{x}} dx \\ \int \frac{\sin^{- 1} x}{\frac{π}{2}} dx [∵ \sin^{- 1} \sqrt{x} + \cos^{- 1} \sqrt{x} = \frac{π}{2}] \\ = \frac{π}{2} \int \sin^{- 1} x . dx [Using Integration by part] \\ = \sin^{- 1} x \int dx - \int [\int dx] d (\sin^{- 1} x) \\ = {xsin}^{- 1} x - \int \frac{xdx}{\sqrt{1 - x^{2}}} \\ Put 1 - x^{2} = t^{2} \\ On differentiating \\ - 2 xdx = 2 tdt \\ \Rightarrow xdx = - tdt \\ = {xsin}^{- 1} x + \int \frac{tdt}{t} \\ = {xsin}^{- 1} x + \int dt \\ = {xsin}^{- 1} x + t + c \\ = {xsin}^{- 1} x + \sqrt{1 - x^{2}} + c \end{array}$

Q.32

$Evaluate \int e^{x} [\sqrt{x} + \frac{1}{2 \sqrt{x}}] dx .$

Ans

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

Q.33

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

Ans

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

Q.34

$Evaluate \int_{0}^{\frac{π}{2}} \sin^{2} x dx .$

Ans

$\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \sin^{2} x dx . .. (1) \\ \Rightarrow I = \int_{0}^{\frac{π}{2}} \sin^{2} (\frac{π}{2} - x) dx \\ \Rightarrow I = \int_{0}^{\frac{π}{2}} \cos^{2} x dx . \dots (2) \\ On adding (1) and (2), \\ 2 I = \int_{0}^{\frac{π}{2}} (\sin^{2} x + \cos^{2} x) dx = \int_{0}^{\frac{π}{2}} 1 . dx \\ \Rightarrow 2 I = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} - 0 \\ \Rightarrow I = \frac{π}{4} \end{array}$

Q.35

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

Ans

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

Q.36

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

Ans

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

Q.37

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

Ans

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

Q.38 Find the anti-derivative of tanx.

Ans

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

Q.39

$Evaluate \int d (sinx) .$

Ans

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

Q.40

$Evaluate \int \log x dx .$

Ans

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

Q.41

$Evaluate \int_{- 1}^{1} logx dx .$

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

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

Ans

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

Q.43

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

Ans

$\begin{array}{l} We have, \int \frac{e^{x - 1} + x^{e - 1}}{e^{x} + x^{e}} dx . \\ Put e^{x} + x^{e} = t \\ On differentiating, \\ (e^{x} + {ex}^{e - 1}) dx = dt \\ \Rightarrow e (e^{x - 1} + x^{e - 1}) dx = dt \\ \Rightarrow (e^{x - 1} + x^{e - 1}) dx = \frac{dt}{e} \\ Now, \frac{1}{e} \int \frac{dt}{t} = \frac{1}{e} \log |t| + c \\ = \frac{1}{e} \log |e^{x} + x^{e}| + c \end{array}$

Q.44

$Evaluate \int_{0}^{1 .5} [x] dx, where [x] is greatest interger function .$

Ans

$\begin{array}{l} We have, \int_{0}^{1 .5} [x] dx = \int_{0}^{1} [x] dx + \int_{0}^{1 .5} [x] dx \\ = \int_{0}^{1} 0 dx + \int_{0}^{1 .5} 1 dx \\ = 0 + {(x)}_{1}^{1 .5} \\ = 1 .5 - 1 \\ = 0 .5 \end{array}$

Q.45

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

Ans

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

Q.46

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

Ans

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

Q.47

$Evaluate \int \sec x dx .$

Ans

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

Q.48

$Evaluate \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{7} x dx .$

Ans

$\begin{array}{l} Let f (x) = \sin^{7} x \\ ∵ f (- x) = \sin^{7} (- x) \\ = - \sin^{7} x \\ = - f (x) \\ ∴ f (x) = \sin^{7} x is an odd function . \\ \Rightarrow \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{7} x dx = 0 \end{array}$

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Practice is essential to attain high scores in Chapter 7 Mathematics. The Class 12 Mathematics chapter 7 notes help students to understand the topics and concepts and retain them for a longer time. Using these notes will be helpful for quick revision on the day of the examination. All formulas, derivations and concepts are very well explained in the Class 12 Mathematics notes chapter 7.

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CBSE Class 12 Maths Revision Notes Chapter 7

Class 12 Mathematics Chapter 7 Notes

NCERT Class 12 Mathematics Chapter 7 Notes: Main Topics

INTRODUCTION:

INDEFINITE INTEGRAL

Standard Formulas:

Geometric Interpretation:

METHODS OF INTEGRATION:

1. Substitution Method:

Some Special Integrals:

Integrals of the form 1:

Integrals of the form 2:

Integrals of form 3:

Integrals of the form 4:

Integrals of the form 5:

Integrals of the form 6:

Integrals of the form 7:

Integrals of form 8:

Integrals of the form 9:

METHOD OF PARTIAL FRACTIONS:

METHOD OF INTEGRATION BY PARTS

Chapter 7 Mathematics Class 12 Notes: Exercises & Solutions

NCERT Exemplar Class 12 Mathematics

Key Features Of Class 12 Mathematics Chapter 7 Notes

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FAQs (Frequently Asked Questions)

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CBSE Class 12 Maths Revision Notes Chapter 7

Class 12 Mathematics Chapter 7 Notes

NCERT Class 12 Mathematics Chapter 7 Notes: Main Topics

INTRODUCTION:

INDEFINITE INTEGRAL

Standard Formulas:

Geometric Interpretation:

METHODS OF INTEGRATION:

1. Substitution Method:

Some Special Integrals:

Integrals of the form 1:

Integrals of the form 2:

Integrals of form 3:

Integrals of the form 4:

Integrals of the form 5:

Integrals of the form 6:

Integrals of the form 7:

Integrals of form 8:

Integrals of the form 9:

METHOD OF PARTIAL FRACTIONS:

METHOD OF INTEGRATION BY PARTS

Chapter 7 Mathematics Class 12 Notes: Exercises & Solutions

NCERT Exemplar Class 12 Mathematics

Key Features Of Class 12 Mathematics Chapter 7 Notes

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FAQs (Frequently Asked Questions)

1. How can I score well using the Class 12 Mathematics chapter 7?

2. How can Extramarks help me in my Chapter 7 Mathematics exam preparation?

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