Self Assessment and Revision on Definite Integrals Class 12 OP Malhotra ISC Maths Solutions Ch-16. In this article you would learn full concept recap of Definite Integrals. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Self Assessment and Revision on Definite Integrals Class 12 OP Malhotra  ISC Maths Solutions Ch-16

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-16	Definite Integrals
Writer	OP Malhotra
Exe	Self Assessment and Revision

Self Assessment and Revision on Definite Integrals

 OP Malhotra ISC Class 12 Maths Solutions

1. Evaluate: ∫ (x³ + 1) / [x(x − 1)] dx

Ans- (x³ + 1)/(x(x−1))
Partial fraction se:
(x³ +1)/(x(x−1)) = x + 1 − 1/x + 2/(x−1)

∫(x + 1 − 1/x + 2/(x−1)) dx
= ∫x dx + ∫1 dx − ∫1/x dx + 2∫1/(x−1) dx
= x²/2 + x − ln|x| + 2ln|x−1| + C

2. Evaluate: ∫ x (log x)² dx

Ans- Using integration by parts
Let u = (log x)² , dv = x dx

du = 2(log x)/x dx
v = x²/2

∫x(logx)² dx = uv − ∫vdu
= (x²/2)(logx)² − ∫(x²/2)(2logx/x) dx
= (x²/2)(logx)² − ∫x logx dx

Again integration by parts:
(x²/2)(logx)² − (x²/2)logx + x²/4 + C

3. Evaluate: ∫ 2/(4x² − 1) dx

Ans- 4x² −1 = (2x−1)(2x+1)

Partial fraction:
2/(4x²−1) = 1/(2x−1) − 1/(2x+1)

∫[1/(2x−1) − 1/(2x+1)] dx
= (1/2)ln|2x−1| − (1/2)ln|2x+1| + C
= (1/2) ln|(2x−1)/(2x+1)| + C

4. Evaluate: ∫ eˣ (log x + 1/x) dx

Ans- Observe
d/dx (eˣ log x) = eˣ log x + eˣ(1/x)
∫ eˣ(log x +1/x) dx = eˣ log x + C

5. Evaluate: ∫ eˣ cos x dx

Ans- Given:
∫ e^x cos x dx

Using formula:
∫ e^x cos x dx = e^x(sin x + cos x)/2 + C

= (e^x/2)(sin x + cos x) + C

6. Evaluate: ∫₀¹ e⁻ˣ/(1 + eˣ) dx

Ans- Given:
∫₀¹ e^−x / (1 + e^x) dx

Multiply numerator & denominator by e^−x:
= ∫₀¹ 1 / (e^x + 1) · e^−x dx
= ∫₀¹ 1 / (1 + e^x) · e^−x dx

Let: t = e^x ⇒ dt = e^x dx

Use symmetry:
I = ∫₀¹ e^−x/(1 + e^x) dx
I = ∫₀¹ 1/(1 + e^x) dx − I
⇒ 2I = ∫₀¹ 1/(1 + e^x) dx

Now integrate:
∫ 1/(1 + e^x) dx = x − ln(1 + e^x)

Apply limits:
2I = [x − ln(1 + e^x)]₀¹
= (1 − ln(1 + e)) − (0 − ln 2)
= 1 − ln(1 + e) + ln 2

I = (1/2) [1 + ln 2 − ln(1 + e)]

7. Evaluate: ∫ log(log x) / x dx

Ans- Given:
∫ log(log x) / x dx

Let:
t = log x ⇒ dt = dx/x
⇒ Integral becomes:
∫ log t dt

Using formula: ∫ log t dt = t log t − t

substitute:
t = log x

= (log x)(log(log x)) − log x + C

8. Evaluate: ∫ₐᵇ (log x)/x dx

Ans- Given:
∫_a^b (log x)/x dx

Let:
t = log x ⇒ dt = dx/x
⇒ Integral becomes:

∫ t dt = t²/2

Apply limits:
= [ (log x)² / 2 ]_a^b

= (1/2) [ (log b)² − (log a)² ]

9. Evaluate: ∫ x sin⁻¹x / √(1−x²) dx

Ans- Given:
∫ x sin⁻¹x / √(1 − x²) dx

Let: t = sin⁻¹x ⇒ x = sin t, dx = cos t dt
⇒ √(1 − x²) = cos t

⇒ Integral becomes:
∫ t · sin t dt

By parts:
= −t cos t + sin t + C

Back substitute:
t = sin⁻¹x, cos t = √(1 − x²), sin t = x

= x − (sin⁻¹x)√(1 − x²) + C

10. Evaluate: ∫₀²π √(1 + sin(x/2)) dx

Ans- Let: t = x/2 ⇒ dx = 2 dt
Limits: 0 → π

⇒ Integral becomes:
2 ∫₀^π √(1 + sin t) dt

Use identity:
1 + sin t = (sin(t/2) + cos(t/2))²
⇒ √(1 + sin t) = sin(t/2) + cos(t/2)
⇒ 2 ∫₀^π [sin(t/2) + cos(t/2)] dt
= 2 [ -2cos(t/2) + 2sin(t/2) ]₀^π
= 2 [ (0 + 2) – (-2 + 0) ]
= 2 × 4 = 8

11. Evaluate: ∫ sin⁻¹(2x/(1+x²)) dx

Ans- Using the identity :
sin⁻¹(2x/(1+x²)) = 2 tan⁻¹x
∫2tan⁻¹x dx

Using integration by parts :
2x tan⁻¹x − ln(1+x²) + C

12. Evaluate: ∫ (x−1)/((x−3)(x−2)²) dx

Ans- Using partial fraction:
(x−1)/((x−3)(x−2)²)
= 2/(x−3) − 2/(x−2) − 1/(x−2)²

Integrating:
2ln|x−3| − 2ln|x−2| + 1/(x−2) + C

13. Evaluate: ∫ eˣ (1 + sin x)/(1 + cos x) dx

Ans- Given:
∫ e^x(1 + sin x)/(1 + cos x) dx

Split:
= ∫ e^x dx + ∫ e^x (sin x)/(1 + cos x) dx

Use:
d/dx[ln(1 + cos x)] = − sin x/(1 + cos x)

⇒ ∫ e^x (sin x)/(1 + cos x) dx
= − ∫ e^x d[ln(1 + cos x)]
= e^x − e^x ln(1 + cos x) + C

14. Prove: ∫₀^(π/2) √(sin x) / (√(sin x) + √(cos x)) dx

Ans- Let I = ∫₀^(π/2) √(sinx)/(√(sinx)+√(cosx)) dx
x → (π/2 − x) substitute this

I = ∫₀^(π/2) √(cosx)/(√(sinx)+√(cosx)) dx

Add both:
2I = ∫₀^(π/2) dx
2I = π/2
I = π/4

16. ∫ e^-2x sin x dx

Ans- Let I = ∫ e^-2x sin x dx

Using formula:
∫ e^ax sin bx dx = e^ax(a sin bx − b cos bx)/(a² + b²) + C

Here a = −2 and b = 1
I = e^-2x(−2 sin x − cos x)/(4 + 1) + C
I = e^-2x(−2 sin x − cos x)/5 + C

17. ∫₀^π/4 (2 cos 2x)/(1 + sin 2x) dx

Ans- Let I = ∫ (2 cos2x)/(1 + sin2x) dx

Put t = 1 + sin2x
dt/dx = 2cos2x

dt = 2cos2x dx
I = ∫ dt/t
I = log|t|
I = log(1 + sin2x)

Applying limits
I = log(1 + sin(π/2)) − log(1 + sin0)
I = log(2) − log(1)
I = log 2

18. ∫₀¹ x tan^-1x dx

Ans- Use integration by parts

Let u = tan^-1x
dv = x dx
du = 1/(1 + x²) dx

v = x²/2

I = uv − ∫v du
I = (x²/2)tan^-1x − ∫ (x²/2)(1/(1 + x²)) dx
x²/(1 + x²) = 1 − 1/(1 + x²)

I = (x²/2)tan^-1x − 1/2 ∫(1 − 1/(1 + x²)) dx
I = (x²/2)tan^-1x − 1/2(x − tan^-1x)

Applying limits from 0 to 1
I = π/4 − 1/2

27. ∫ 1/[x cos²(1 + log x)] dx

Ans- Let t = 1 + log x
dt/dx = 1/x
dt = dx/x

Integral becomes
I = ∫ sec² t dt
I = tan t + C
I = tan(1 + log x) + C

36. ∫ (x² − 5x − 1)/(x⁴ + x² + 1) dx

Ans- Let I = ∫ (x² − 5x − 1)/(x⁴ + x² + 1) dx

Factor denominator: x⁴ + x² + 1 = (x² − x +1)(x² + x +1)

Resolve into partial fractions
I = A/(x² − x +1) + B/(x² + x +1)

Integrating gives logarithmic terms
I = (1/2) log(x² − x +1) − (1/2) log(x² + x +1) + C

37. ∫₁² √x / √(3 − x + √x) dx

Ans- Given:
∫ √x / √(3 − x + √x) dx

Let: √x = t ⇒ x = t², dx = 2t dt
⇒ Integral becomes:
∫ (t / √(3 − t² + t)) · 2t dt
= ∫ 2t² / √(3 − t² + t) dt
= (15/4) sin⁻¹((√x − 1/2)/(√13/2))− (√x + 3/2) √(3 − x + √x) + C

38. ∫ 1 / [x(6(log x)² + 7 log x + 2)] dx

Ans- Let: t = log x ⇒ dt = dx/x
⇒ Integral becomes:
∫ 1 / (6t² + 7t + 2) dt

Factorize:
6t² + 7t + 2 = (3t + 2)(2t + 1)
⇒ ∫ [1/(3t+2)(2t+1)] dt

By partial fractions:
= ∫ [3/(3t+2) − 2/(2t+1)] dt
= ln|3log x + 2| − ln|2log x + 1| + C

39. ∫ 1/(x + √x) dx

Ans- Let √x = t
x = t²
dx = 2t dt

Substitute in the integral|
I = ∫ 1/(t² + t) · 2t dt
I = ∫ 2t/(t(t+1)) dt
I = ∫ 2/(t+1) dt
I = 2 log|t+1| + C

Substitute t = √x
I = 2 log(√x + 1) + C

45. ∫ 2y²/(y² + 4) dy

Ans- Rewrite the numerator
2y² = 2(y² + 4) − 8
I = ∫ [2(y²+4) − 8]/(y² + 4) dy
I = ∫ 2 dy − ∫ 8/(y² + 4) dy
I = 2y − 8 ∫ 1/(y² + 4) dy

Use formula ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a)
I = 2y − 8 × (1/2) tan⁻¹(y/2) + C
I = 2y − 4 tan⁻¹(y/2) + C

48. ∫ tan³x dx

Ans- tan³x = tanx(tan²x)
tan²x = sec²x − 1
tan³x = tanx(sec²x − 1)

I = ∫ tanx sec²x dx − ∫ tanx dx

For first integral let t = secx
dt = secx tanx dx
∫ tanx sec²x dx = sec²x /2

Now ∫ tanx dx = −log|cosx|
I = sec²x/2 + log|cosx| + C

49. ∫₀^(π/2) √sinx /(√sinx + √cosx) dx

Ans- Let I = ∫₀^(π/2) √sinx /(√sinx + √cosx) dx

Replace x by (π/2 − x)
I = ∫₀^(π/2) √cosx /(√sinx + √cosx) dx

Add both integrals
2I = ∫₀^(π/2) (√sinx + √cosx)/(√sinx + √cosx) dx
2I = ∫₀^(π/2) 1 dx
2I = π/2
I = π/4

50. ∫₀^(π/2) (sinx − cosx)/(1 + sinx cosx) dx

Ans-Let I = ∫₀^(π/2) (sinx − cosx)/(1 + sinx cosx) dx
Replace x by (π/2 − x)

I = ∫₀^(π/2) (cosx − sinx)/(1 + sinx cosx) dx

Add both integrals
2I = ∫₀^(π/2) 0 dx
2I = 0
I = 0

–: End of Self Assessment and Revision on Definite Integrals Class 12 OP Malhotra ISC Maths Solutions :–

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