Class 12 Maths Chapter 7 Important Questions – Integrals

Integration is the inverse process of differentiation and is used to find antiderivatives, definite values and areas associated with functions. The chapter covers indefinite and definite integrals, substitution, partial fractions, integration by parts, the Fundamental Theorem of Calculus and properties of definite integrals.

Class 12 Maths Chapter 7 Important Questions help students practise the main methods used to evaluate integrals. The questions cover standard formulas, substitution, trigonometric identities, partial fractions, integration by parts and definite integrals.

Students should first identify the form of the integrand before choosing a method. An expression containing a function and its derivative generally requires substitution, while a product involving logarithmic, inverse trigonometric, algebraic or exponential functions may require integration by parts.

Key Takeaways

• Integration is the inverse process of differentiation.
• An indefinite integral contains an arbitrary constant C.
• A definite integral has a fixed numerical value.
• Substitution is useful when the derivative of an inner function is present.
• Partial fractions are used for proper rational functions.
• Integration by parts is useful for products of functions.
• Properties of definite integrals can reduce lengthy calculations.

Important Integrals Formulas

Power formula:

∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, where n ≠ −1

∫ dx = x + C

∫ 1/x dx = log|x| + C

Exponential functions:

∫ eˣ dx = eˣ + C

∫ aˣ dx = aˣ/log a + C

Trigonometric functions:

∫ sin x dx = −cos x + C

∫ cos x dx = sin x + C

∫ sec²x dx = tan x + C

∫ cosec²x dx = −cot x + C

∫ sec x tan x dx = sec x + C

∫ cosec x cot x dx = −cosec x + C

∫ tan x dx = log|sec x| + C

∫ cot x dx = log|sin x| + C

∫ sec x dx = log|sec x + tan x| + C

∫ cosec x dx = log|cosec x − cot x| + C

Inverse trigonometric forms:

∫ dx/(1 + x²) = tan⁻¹x + C

∫ dx/√(1 − x²) = sin⁻¹x + C

∫ dx/(x² + a²) = (1/a)tan⁻¹(x/a) + C

∫ dx/√(a² − x²) = sin⁻¹(x/a) + C

Integration by parts:

∫ u dv = uv − ∫ v du

Fundamental Theorem of Calculus:

If F′(x) = f(x), then:

∫ from a to b f(x) dx = F(b) − F(a)

Access Class 12 Maths Chapter 7 Important Questions in 30 Minutes

• First 10 minutes: Revise standard integrals and integration by inspection.
• Next 10 minutes: Practise substitution, trigonometric identities and partial fractions.
• Final 10 minutes: Solve integration-by-parts and definite-integral questions.

Need more practice with indefinite and definite integrals?

Practise chapter-wise questions and review step-by-step solutions on the Extramarks Learning App. Sign Up Free

Multiple Choice and Very Short Answer Questions – 1 Mark

Q1. What is integration?

Answer: Integration is the inverse process of differentiation. It is also called antidifferentiation.

Q2. What is an antiderivative of a function f(x)?

Answer: A function F(x) is an antiderivative of f(x) if:

F′(x) = f(x)

Q3. Why is a constant C added to an indefinite integral?

Answer: The derivative of every constant is zero. Therefore, all functions of the form F(x) + C have the same derivative.

Q4. Evaluate ∫ dx.

Answer:

∫ dx = x + C

Q5. Evaluate ∫ x⁵ dx.

Answer:

∫ x⁵ dx = x⁶/6 + C

Q6. Evaluate ∫ cos x dx.

Answer:

∫ cos x dx = sin x + C

Q7. Evaluate ∫ sec²x dx.

Answer:

∫ sec²x dx = tan x + C

Q8. Evaluate ∫ 1/x dx.

Answer:

∫ 1/x dx = log|x| + C

Q9. Evaluate ∫ eˣ dx.

Answer:

∫ eˣ dx = eˣ + C

Q10. Evaluate ∫ dx/(1 + x²).

Answer:

∫ dx/(1 + x²) = tan⁻¹x + C

Q11. What is the difference between indefinite and definite integrals?

Answer: An indefinite integral represents a family of antiderivatives and contains C. A definite integral has fixed limits and gives a numerical value.

Q12. State the value of ∫ from a to a f(x) dx.

Answer:

∫ from a to a f(x) dx = 0

Short Answer Questions – 2 or 3 Marks

Q13. Find ∫ (4x³ − 6) dx.

Solution:

Integrate each term separately:

∫ (4x³ − 6) dx

= 4∫x³ dx − 6∫dx

= 4(x⁴/4) − 6x + C

= x⁴ − 6x + C

Q14. Find the antiderivative F of f(x) = 4x³ − 6 if F(0) = 3.

Solution:

∫ (4x³ − 6) dx = x⁴ − 6x + C

Therefore:

F(x) = x⁴ − 6x + C

Given:

F(0) = 3

3 = 0 − 0 + C

C = 3

Hence:

F(x) = x⁴ − 6x + 3

Q15. Evaluate ∫ (x³ − 1)/x² dx.

Solution:

First simplify:

(x³ − 1)/x² = x − 1/x²

Therefore:

∫ (x³ − 1)/x² dx

= ∫ (x − x⁻²) dx

= x²/2 + 1/x + C

Q16. Evaluate ∫ (1 − x)√x dx.

Solution:

(1 − x)√x = x¹ᐟ² − x³ᐟ²

Therefore:

∫ (1 − x)√x dx

= ∫ x¹ᐟ² dx − ∫ x³ᐟ² dx

= (2/3)x³ᐟ² − (2/5)x⁵ᐟ² + C

Q17. Evaluate ∫ sec x(sec x + tan x) dx.

Solution:

Expand the integrand:

∫ sec x(sec x + tan x) dx

= ∫ (sec²x + sec x tan x) dx

= tan x + sec x + C

Q18. Evaluate ∫ (2 − 3 sin x)/cos²x dx.

Solution:

Rewrite the integrand:

(2 − 3 sin x)/cos²x

= 2 sec²x − 3 sec x tan x

Therefore:

∫ (2 − 3 sin x)/cos²x dx

= 2∫sec²x dx − 3∫sec x tan x dx

= 2 tan x − 3 sec x + C

Q19. Evaluate ∫ (1 − sin x)/cos²x dx.

Solution:

Rewrite:

(1 − sin x)/cos²x

= sec²x − sec x tan x

Therefore:

∫ (1 − sin x)/cos²x dx

= tan x − sec x + C

Q20. Evaluate ∫ (log x)/x dx.

Solution:

Let:

t = log x

Then:

dt = dx/x

Therefore:

∫ (log x)/x dx = ∫ t dt

= t²/2 + C

Substituting t = log x:

∫ (log x)/x dx = (log x)²/2 + C

Q21. Evaluate ∫ e^(tan⁻¹x)/(1 + x²) dx.

Solution:

Let:

t = tan⁻¹x

Then:

dt = dx/(1 + x²)

Therefore:

∫ e^(tan⁻¹x)/(1 + x²) dx

= ∫ eᵗ dt

= eᵗ + C

Substituting:

= e^(tan⁻¹x) + C

Integration by Substitution – 3 or 4 Marks

Q22. Evaluate ∫ 2x sin(x² + 1) dx.

Solution:

Let:

t = x² + 1

Then:

dt = 2x dx

Therefore:

∫ 2x sin(x² + 1) dx

= ∫ sin t dt

= −cos t + C

Substituting:

= −cos(x² + 1) + C

Q23. Evaluate ∫ x/(x² + 4) dx.

Solution:

Let:

t = x² + 4

Then:

dt = 2x dx

x dx = dt/2

Therefore:

∫ x/(x² + 4) dx

= 1/2 ∫ dt/t

= 1/2 log|t| + C

= 1/2 log(x² + 4) + C

Q24. Evaluate ∫ cos x/(1 + sin x) dx.

Solution:

Let:

t = 1 + sin x

Then:

dt = cos x dx

Therefore:

∫ cos x/(1 + sin x) dx

= ∫ dt/t

= log|t| + C

= log|1 + sin x| + C

Q25. Evaluate ∫ x²e^(x³) dx.

Solution:

Let:

t = x³

Then:

dt = 3x² dx

x² dx = dt/3

Therefore:

∫ x²e^(x³) dx

= 1/3 ∫ eᵗ dt

= eᵗ/3 + C

= e^(x³)/3 + C

Q26. Evaluate ∫ tan²x sec²x/√(tan²x + 4) dx.

Solution:

Let:

t = tan x

Then:

dt = sec²x dx

The integral becomes:

∫ t²/√(t² + 4) dt

Write:

t² = (t² + 4) − 4

Therefore:

∫ t²/√(t² + 4) dt

= ∫ √(t² + 4) dt − 4∫ dt/√(t² + 4)

Using standard integrals:

= 1/2[t√(t² + 4) + 4log|t + √(t² + 4)|]

− 4log|t + √(t² + 4)| + C

= 1/2 t√(t² + 4) − 2log|t + √(t² + 4)| + C

Substituting t = tan x:

= 1/2 tan x√(tan²x + 4)

− 2log|tan x + √(tan²x + 4)| + C

Integration Using Trigonometric Identities – 3 or 4 Marks

Q27. Evaluate ∫ cos²x dx.

Solution:

Use:

cos²x = (1 + cos 2x)/2

Therefore:

∫ cos²x dx

= 1/2 ∫ dx + 1/2 ∫ cos 2x dx

= x/2 + sin 2x/4 + C

Q28. Evaluate ∫ sin 2x cos 3x dx.

Solution:

Use:

sin A cos B = 1/2[sin(A + B) + sin(A − B)]

Therefore:

sin 2x cos 3x

= 1/2[sin 5x − sin x]

Hence:

∫ sin 2x cos 3x dx

= 1/2∫sin 5x dx − 1/2∫sin x dx

= −cos 5x/10 + cos x/2 + C

Q29. Evaluate ∫ sin³x dx.

Solution:

Write:

sin³x = sin x(1 − cos²x)

Therefore:

∫ sin³x dx

= ∫ sin x(1 − cos²x) dx

Let:

t = cos x

dt = −sin x dx

Then:

= −∫(1 − t²) dt

= −t + t³/3 + C

Substituting:

= −cos x + cos³x/3 + C

Q30. Evaluate ∫ dx/(sin²x cos²x).

Solution:

Use:

sin²x + cos²x = 1

Therefore:

1/(sin²x cos²x)

= (sin²x + cos²x)/(sin²x cos²x)

= sec²x + cosec²x

Hence:

∫ dx/(sin²x cos²x)

= ∫sec²x dx + ∫cosec²x dx

= tan x − cot x + C

Integration by Partial Fractions – 4 Marks

Q31. Evaluate ∫ dx/[(x + 1)(x + 2)].

Solution:

Let:

1/[(x + 1)(x + 2)]

= A/(x + 1) + B/(x + 2)

Therefore:

1 = A(x + 2) + B(x + 1)

Putting x = −1:

1 = A

Putting x = −2:

1 = −B

Therefore:

A = 1 and B = −1

Hence:

∫ dx/[(x + 1)(x + 2)]

= ∫[1/(x + 1) − 1/(x + 2)] dx

= log|x + 1| − log|x + 2| + C

= log|(x + 1)/(x + 2)| + C

Q32. Evaluate ∫ x/[(x − 1)(x − 2)] dx.

Solution:

Let:

x/[(x − 1)(x − 2)]

= A/(x − 1) + B/(x − 2)

Then:

x = A(x − 2) + B(x − 1)

Putting x = 1:

1 = −A

A = −1

Putting x = 2:

2 = B

Therefore:

x/[(x − 1)(x − 2)]

= −1/(x − 1) + 2/(x − 2)

Hence:

∫ x/[(x − 1)(x − 2)] dx

= −log|x − 1| + 2log|x − 2| + C

Q33. Evaluate ∫ (x² + x + 1)/[(x + 2)(x² + 1)] dx.

Solution:

Write:

(x² + x + 1)/[(x + 2)(x² + 1)]

= A/(x + 2) + (Bx + C)/(x² + 1)

Therefore:

x² + x + 1

= A(x² + 1) + (Bx + C)(x + 2)

Comparing coefficients gives:

A = 3/5

B = 2/5

C = 1/5

Therefore:

∫ (x² + x + 1)/[(x + 2)(x² + 1)] dx

= 3/5∫dx/(x + 2)

• 1/5∫(2x)/(x² + 1) dx
• 1/5∫dx/(x² + 1)

Hence:

= 3/5 log|x + 2|

• 1/5 log(x² + 1)
• 1/5 tan⁻¹x + C

Integration by Parts – 4 Marks

Q34. Evaluate ∫ x cos x dx.

Solution:

Using integration by parts:

∫ u dv = uv − ∫v du

Let:

u = x

dv = cos x dx

Then:

du = dx

v = sin x

Therefore:

∫ x cos x dx

= x sin x − ∫sin x dx

= x sin x + cos x + C

Q35. Evaluate ∫ x eˣ dx.

Solution:

Let:

u = x

dv = eˣ dx

Then:

du = dx

v = eˣ

Therefore:

∫ x eˣ dx

= xeˣ − ∫eˣ dx

= xeˣ − eˣ + C

= eˣ(x − 1) + C

Q36. Evaluate ∫ log x dx.

Solution:

Write:

∫ log x dx = ∫ log x · 1 dx

Let:

u = log x

dv = dx

Then:

du = dx/x

v = x

Therefore:

∫ log x dx

= x log x − ∫x(1/x) dx

= x log x − ∫dx

= x log x − x + C

Q37. Evaluate ∫ sin⁻¹x dx.

Solution:

Let:

u = sin⁻¹x

dv = dx

Then:

du = dx/√(1 − x²)

v = x

Therefore:

∫ sin⁻¹x dx

= x sin⁻¹x − ∫x/√(1 − x²) dx

Let:

t = 1 − x²

dt = −2x dx

Then:

∫x/√(1 − x²) dx = −√(1 − x²)

Therefore:

∫ sin⁻¹x dx

= x sin⁻¹x + √(1 − x²) + C

Q38. Evaluate ∫ eˣ sin x dx.

Solution:

Let:

I = ∫eˣ sin x dx

Using integration by parts:

I = eˣ sin x − ∫eˣ cos x dx

Let:

J = ∫eˣ cos x dx

Again using integration by parts:

J = eˣ cos x + ∫eˣ sin x dx

J = eˣ cos x + I

Therefore:

I = eˣ sin x − eˣ cos x − I

2I = eˣ(sin x − cos x)

Hence:

I = eˣ(sin x − cos x)/2 + C

Definite Integrals – 3 or 4 Marks

Q39. Evaluate ∫ from 0 to 1 x² dx.

Solution:

∫ x² dx = x³/3

Therefore:

∫ from 0 to 1 x² dx

= [x³/3] from 0 to 1

= 1/3 − 0

= 1/3

Q40. Evaluate ∫ from 0 to π/2 sin x dx.

Solution:

∫ sin x dx = −cos x

Therefore:

∫ from 0 to π/2 sin x dx

= [−cos x] from 0 to π/2

= −cos(π/2) + cos 0

= 0 + 1

= 1

Q41. Evaluate ∫ from 0 to 1 x/(1 + x²) dx.

Solution:

Let:

t = 1 + x²

dt = 2x dx

Therefore:

∫ from 0 to 1 x/(1 + x²) dx

= 1/2 ∫ from 1 to 2 dt/t

= 1/2[log t] from 1 to 2

= 1/2(log 2 − log 1)

= 1/2 log 2

Q42. Evaluate ∫ from 0 to π/4 tan x dx.

Solution:

∫ tan x dx = log|sec x|

Therefore:

∫ from 0 to π/4 tan x dx

= [log|sec x|] from 0 to π/4

= log√2 − log 1

= log√2

= 1/2 log 2

Q43. If f(x) = ∫ from 0 to x t sin t dt, find f′(x).

Solution:

By the first Fundamental Theorem of Calculus:

If:

f(x) = ∫ from 0 to x g(t) dt

then:

f′(x) = g(x)

Here:

g(t) = t sin t

Therefore:

f′(x) = x sin x

Properties of Definite Integrals – 4 or 5 Marks

Q44. State the property for reversing the limits of a definite integral.

Answer:

∫ from a to b f(x) dx = −∫ from b to a f(x) dx

Q45. State the property involving the substitution x = a + b − t.

Answer:

∫ from a to b f(x) dx

= ∫ from a to b f(a + b − x) dx

Q46. Evaluate ∫ from −a to a f(x) dx when f is an odd function.

Answer:

If f(−x) = −f(x), then:

∫ from −a to a f(x) dx = 0

Q47. Evaluate ∫ from −a to a f(x) dx when f is an even function.

Answer:

If f(−x) = f(x), then:

∫ from −a to a f(x) dx

= 2∫ from 0 to a f(x) dx

Q48. Evaluate ∫ from 0 to π/2 sin²x dx.

Solution:

Let:

I = ∫ from 0 to π/2 sin²x dx

Using the property:

I = ∫ from 0 to π/2 cos²x dx

Adding:

2I = ∫ from 0 to π/2 (sin²x + cos²x) dx

2I = ∫ from 0 to π/2 dx

2I = π/2

Therefore:

I = π/4

Q49. Evaluate ∫ from 0 to π/2 sin x/(sin x + cos x) dx.

Solution:

Let:

I = ∫ from 0 to π/2 sin x/(sin x + cos x) dx

Using the property x → π/2 − x:

I = ∫ from 0 to π/2 cos x/(sin x + cos x) dx

Adding the two equations:

2I = ∫ from 0 to π/2 1 dx

2I = π/2

Therefore:

I = π/4

Practice Questions for Class 12 Maths Chapter 7

1. Evaluate ∫ (3x² + 4x³) dx.
2. Evaluate ∫ (2x + 1)e^(x² + x) dx.
3. Evaluate ∫ sin x/(1 + cos x) dx.
4. Evaluate ∫ sin²x dx.
5. Evaluate ∫ sin 3x cos 4x dx.
6. Evaluate ∫ dx/(x² − 9).
7. Evaluate ∫ (3x + 5)/(x² + 3x − 18) dx.
8. Evaluate ∫ x log x dx.
9. Evaluate ∫ x²eˣ dx.
10. Evaluate ∫ tan⁻¹x dx.
11. Evaluate ∫ from 0 to 3 dx/(9 + x²).
12. Evaluate ∫ from 0 to π/4 sin 2x dx.
13. Evaluate ∫ from 0 to 1 2x/(1 + x²) dx.
14. Evaluate ∫ from −2 to 2 (x³ + x) dx.
15. Evaluate ∫ from 0 to π/2 cos x/(sin x + cos x) dx.

Important Questions Class 12 Maths Chapter-Wise