Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.10
![Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.10 Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.10](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6cS96WBBtqQ6dpB-m0R_6OmT7N7SCieY4RScqrQPJUZrYZRbx5rIxir3i0bG6gxt6Ox1NZnkPgvt9Tc7y0AcFwp-jCHUe_pK7Tz1C8o4GUPWE_B32y0dUFSnKbaxaB_3RXFN2xDqw6zeLKUgJIxITMxsoDP2JMF2pHguysCaWU-WysDDILYbyN8IY/w663-h313-rw/cbse-class-12-maths-ncert-solutions-chapter-7-exercise-7-10.jpg)
Integrals Exercise 7.10 Solutions
1. Evaluate the integrals in using substitution
∫01 x/(x2 + 1) dx
Solution
Let x2 + 1 = t ⇒ 2x dx = dt
when x = 0, t = 1 and when x = 1, t = 2
2. Evaluate the integrals in using substitution
∫0Ï€/2 √(sin Ø) cos5Ñ„ dÑ„
Solution
= 64/231
3. ∫01 sin-1 [2x/(1 + x2)] dx
Solution
Consider, I = ∫01 sin-1 [2x/(1 + x2 )] dx
Let x = tan θ ⇒ dx = sec2 θ dθ
When x = 0, θ = 0 and when x = 1, θ = π/4
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCzfM_LzKXHe4k0-Ji-KHKfXfyXt_wJNke4G3xRRjq7TzASgkk3hgs5C79d7b1SLwot1y3eN8I0tCDRxAKZP1Q_WEhSar7Q5r3GHDv-89DhkGGYd40EcXZ30e6GRRli_8007TPKW-BS0Fhv-IaW3KynMFk-SnGwQ9TtJ1JJal3ToSazGzoKWVr8yig/s1600-rw/NCERT%20Solutions%20for%20Chapter%207%20Integrals%20Class%2012%20Maths%20Exercise%20-%207.10%20img%204.JPG)
Taking θ as first function and sec2 θ as second function and integrating by parts, we obtain
When x = 0, θ = 0 and when x = 1, θ = π/4
Taking θ as first function and sec2 θ as second function and integrating by parts, we obtain
4. Evaluate the integrals in using substitution ∫02 [x√(x + 2)] (Put x + 2 = t2)
Solution
∫02 [x√(x + 2)]
Let x + 2 = t2 ⇒ dx = 2tdt
When x = 0, t = √2 and when x = 2, t = 2
Let x + 2 = t2 ⇒ dx = 2tdt
When x = 0, t = √2 and when x = 2, t = 2
5. Evaluate the integrals in using substitution ∫0Ï€/2 sin x/(1 + cos2 x) dx
Solution
∫0Ï€/2 sin x/(1 + cos2 x) dx
Let cos x = t ⇒ -sin x dx = dt
when x = 0, t = 1 and when x = π/2, t = 0
Let cos x = t ⇒ -sin x dx = dt
when x = 0, t = 1 and when x = π/2, t = 0
6. Evaluate the integrals in using substitution ∫02 dx/(x + 4 - x2)
Solution
Let x - 1/2 = t ⇒ dx = dt
When x = 0, t = -1/2 and when x = 2, t = 3/2
7. Evaluate the integrals in using substitution ∫-11 dx/(x2 + 2x + 5)
Solution
Let x + 1 = t ⇒ dx = dt
When x = -1, t = 0 and when x = 1, t = 2
8. Evaluate the integrals in using substitution ∫12 (1/x - 1/2x2) e2x dx
Solution
Let 2x = t ⇒ 2dx = dt
When x = 1, t = 2 and when x = 2, t = 4
9. The value of the integral ∫1/34 [(x - x3 )1/3]/x4 dx is
(A) 6
(B) 0
(C) 3
(D) 4
(B) 0
(C) 3
(D) 4
Solution
Let cot θ = t ⇒ - cosec2 θ dθ = dt
When θ = sin-1 (1/3) . t = 2√2 and when θ = Ï€/2 . t = 0
Hence, the correct answer is A.
10. If f(x) = ∫0x t sin t dt, then f'(x) is
(A) cos x + x sin x
(A) cos x + x sin x
(B) x sin x
(C) x cos x
(D) sin x + x cos x
Solution
f(x) = ∫0x t sin t dt
Integrating by parts, we obtain
= - x cos x + sin x
⇒ f'(x) = -[{x(-sin x)} + cos x] + cos x
= x sin x - cos x + cos x
= x sin x
Hence, the correct answer is B.