RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 7 Integrals Ex 7.10

Question 1.
\(\int_{0}^{1} \frac{x}{x^{2}+1}\) dx
Answer:
Let I = \(\int_{0}^{1} \frac{x}{x^{2}+1}\) dx
Put x² + 1 = t
⇒ 2x dx = dt ⇒ x dx = \(\frac{1}{2}\) dt
When x = 0, then t = 1,
When x = 1, then t = 1² + 1 = 2

Question 2.
\(\int_{0}^{\pi / 2} \sqrt{\sin \phi} \cos ^{5} \phi d \phi\)
Answer:
Let I = \(\int_{0}^{\pi / 2} \sqrt{\sin \phi} \cos ^{5} \phi d \phi\)

Question 3.
\(\int_{0}^{1}\) sin^-1 \(\left(\frac{2 x}{1+x^{2}}\right)\) dx
Answer:
Let I = \(\int_{0}^{1}\) sin^-1 \(\left(\frac{2 x}{1+x^{2}}\right)\) dx
Putting x = tan θ ⇒ dx = sec² θ dθ
When x = 1, then θ = \(\frac{\pi}{4}\)
When x = 0, then θ = 0
[Since, tan θ = 1 = tan \(\frac{\pi}{4}\) ∴ θ = \(\frac{\pi}{4}\)]

Now, ∫ θ sec² θ - ∫ {\(\frac{d}{d \theta}\) (θ) ∫sec² θ dθ} dθ
= θ tan θ - ∫1.tan θ dθ
= θ tan θ - (- log |cos θ|)
= θ tan θ + log |cos θ|
∴ ∫ θ sec² θ dθ = θ tan θ + log |cos θ|
Putting the value of ∫θ sec² θ dθ in (i), we get

Question 4.
\(\int_{0}^{2} x\sqrt{x+2}\) dx
Answer:
Let I = \(\int_{0}^{2} x\sqrt{x+2}\) dx
Putting x + 2 = t² ⇒ dx = 2t dt
When x = 0, then t = √2
When x = 2, then t² = 2 + 2 = 4 ⇒ t = 2

Question 5.
\(\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos ^{2} x}\) dx
Answer:
Let I = \(\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos ^{2} x}\) dx,
Putting, cos x = t ⇒ - sin x dx = dt,
⇒ sin x dx = - dt
When x = 0, then t = cos θ = 1,
When x = \(\frac{\pi}{2}\), then t = cos \(\frac{\pi}{2}\) = 0

Question 6.
\(\int_{0}^{2} \frac{d x}{x+4-x^{2}}\)
Answer:

Question 7.
\(\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}\)
Answer:

Question 8.
\(\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x}\) dx
Answer:

Question 9.
The value of the integral
\(\int_{1 / 3}^{1} \frac{\left(x-x^{3}\right)^{1 / 3}}{x^{4}}\) dx is:
(A) 6
(B) 0
(C) 3
(D) 4
Answer:

Again putting cot θ = t ⇒ - cosec² θ dθ = dt
When θ = sin^-1 \(\frac{1}{3}\) ⇒ sin θ = \(\frac{1}{3}\),
Then cot θ = 2√2
∴ t = 2√2 = √8


Hence, (A) is the correct answer.

Question 10.
If f(x) = \(\int_{0}^{x}\) t sin t dt, then f'(x) is:
(A) cot x + x sin x
(B) x sin x
(C) x cos x
(D) sin x + x cos x
Answer:

= - x cos x + sin x
∴ f'(x) = - 1.cos x - x(- sin x) + cos x
= - cos x+ x sin x + cos x
= x sin x
Hence, (B) is the correct answer.