RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 7 Integrals Ex 7.2

Question 1.
\(\frac{2 x}{1+x^{2}}\)
Answer:
Let I = ∫ \(\frac{2 x}{1+x^{2}}\) dx
Putting 1 + x² = t
⇒ 2x dx = dt
∴ ∫ \(\frac{2 x}{1+x^{2}}\) dx = ∫ \(\frac{d t}{t}\)
= log |t| + C
= log |1 + x²| + C
= log (1 + x²) + C

Question 2.
\(\frac{(\log x)^{2}}{x}\)
Answer:

Question 3.
\(\frac{1}{(x+x \log x)}\)
Answer:

Question 4.
sin x sin (cos x)
Answer:
Let I = ∫sin x sin (cos x) dx
Putting cos x = t
⇒ - sin x dx = dt
⇒ sin x dx = - dt
∴ ∫ sin x sin (cos x) dx = - ∫ sin t dt
= + cos t + C
= cos (cos x) + C

Question 5.
sin (ax + b) cos (ax + b)
Answer:
Let I = ∫ sin (ax + b) cos (ax + b) dx
Putting sin (ax + b) = t
⇒ a cos (ax + b) dx = dt
⇒ cos (ax + b) dx = \(\frac{d t}{a}\)

Remark : Here, we can also evaluate integral by assuming cos (ax + b) = t.
Putting cos(ax + b) = t
⇒ - a sin(ax + b) dx = dt
⇒ sin (ax + b) dx = - \(\frac{d t}{a}\)
∴ ∫sin (ax + b) cos(ax + b)dx
= - \(\int \frac{t d t}{a}=-\frac{1}{a} \int t\) dt
= - \(\frac{1}{a} \cdot \frac{t^{2}}{2}\) + C
= - \(\frac{\cos ^{2}(a x+b)}{2 a}\) + C

Second Method:
∫ sin (ax + b) cos (ax + b) dx
= \(\frac{1}{2}\) ∫ 2 sin (ax + b) cos (ax + b) dx
[Multiplying numerator and denominator by 2]
= \(\frac{1}{2}\) ∫ sin 2(ax + b) dx
= \(\frac{1}{2}\) ∫ sin (2ax + 2b) dx
Putting 2ax + 2b = t
⇒ 2a dx = dt

Question 6.
x\(\sqrt{a x+b}\)
Answer:
Let I = ∫ \(\sqrt{a x+b}\) dx = ∫ (ax + b)^1/2 dx
Putting ax + b = t
⇒ a dx = dt
⇒ dx = \(\frac{d t}{a}\)

Question 7.
x\(\sqrt{x+2}\)
Answer:

Question 8.
x\(\sqrt{1+2 x^{2}}\)
Answer:
Let I = ∫ x\(\sqrt{1+2 x^{2}}\) dx
Putting 1 + 2x² = t
⇒ 4 dx = dt
⇒ x dx = \(\frac{d t}{4}\)

Question 9.
(4x + 2) \(\sqrt{x^{2}+x+1}\)
Answer:

Question 10.
\(\frac{1}{x-\sqrt{x}}\)
Answer:

Question 11.
\(\frac{x}{\sqrt{x+4}}\), x > 0
Answer:
Let I = ∫ \(\frac{x}{\sqrt{x+4}}\) dx
Putting x + 4 = t
⇒ dx = dt
and x = t - 4

Question 12.
(x³ - 1)^1/3 x⁵
Answer:
Let I = ∫ (x³ - 1)^1/3 x⁵
Putting x³ - 1 = t
⇒ 3x² dx = dt
⇒ x² dx = \(\frac{d t}{3}\)
and x³ = t + 1
∴ ∫ (x³ - 1)^1/3 x⁵ dx = ∫ (x³ - )^1/3 . x³ .x² dx

Question 13.
\(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\)
Answer:
Let I = ∫ \(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\) dx
Putting 2 + 3x³ = t
⇒ 9x² dx = dt
⇒ x² dx = \(\frac{1}{9}\) dt

Question 14.
\(\frac{1}{x(\log x)^{m}}\), x > 0 , m ≠ 1
Answer:

Question 15.
\(\frac{x}{9-4 x^{2}}\)
Answer:
Let I = ∫ \(\frac{x}{9-4 x^{2}}\) dx
Putting 9 - 4x² = t
⇒ - 8x dx = dt
⇒ x dx = - \(\frac{1}{8} \)dt

Question 16.
e^{2x + 3}
Answer:
Let I = ∫ e^{2x + 3} dx
= ∫ e^2x . e³ dx
= e³ ∫ e^2x dx
Putting 2x = t
⇒ 2 dx = dt
⇒ dx = \(\frac{d t}{2}\)
∴ ∫ e^{2x + 3} dx = e³ ∫ e^t

Alter: ∫ e^{2x + 3} dx
Let 2x + 3 = t
2 dx = dt
or dx = \(\frac{d t}{2}\)
∴ ∫ e^{2x + 3} dx = \(\frac{1}{2}\) ∫ e^t dt = \(\frac{1}{2}\) e^t + C
= \(\frac{1}{2}\) e^{2x + 3} + C

Question 17.
\(\frac{x}{e^{x^{2}}}\)
Answer:

Question 18.
\(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\)
Answer:

Question 19.
\(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Answer:

Question 20.
\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Answer:
Let I = \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Putting e^2x + e^-2x = t
⇒ (2e^2x - 2e^-2x) dx = dt
⇒ (e^2x - e^-2x) = \(\frac{1}{2}\) dt
∴ ∫\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx = \(\frac{1}{2} \int \frac{1}{t}\) dt
= log |t| + C
= log |e^2x + e^-2x| + C

Question 21.
tan² (2x - 3)
Answer:
Let I = ∫ tan² (2x - 3) dx
= ∫ {sec² (2x - 3) - 1} dx
= ∫ sec² (2x - 3) dx - ∫ dx
Putting 2x - 3 = t
⇒ 2dx = dt
⇒ dx = \(\frac{1}{2}\)dt
∴ ∫ tan² (2x - 3) dx = ∫ sec² (2x - 3) dx = ∫ dx
= \(\frac{1}{2}\) ∫ sec² t.dt - ∫ dx
= \(\frac{1}{2}\) tan t - x + C
= \(\frac{1}{2}\) tan (2x - 3) - x + C

Question 22.
sec² (7 - 4x)
Answer:
Let I = ∫ sec² (7 - 4x) dx
Putting 7 - 4x = t
⇒ - 4 dx = dt
⇒ dx = - \(\frac{1}{4}\) dt
∴ ∫ sec² (7 - 4x) dx = - \(\frac{1}{4}\) ∫sec² t dt
= - \(\frac{1}{4}\) tan t + C
= - \(\frac{1}{4}\) tan (7 - 4x) + C

Question 23.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
Answer:
Let I = ∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) dx
Putting sin^-1 x = t

Question 24.
\(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Answer:

Putting 3 cos x + 2 sin x = t
⇒ (- 3 sin x + 2 cos x) dx = dt
⇒ (2 cos x - 3 sin x) dx = dt

Question 25.
Answer:

Question 26.
\(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Answer:

Question 27.
\(\sqrt{\sin 2 x}\) cos 2x
Answer:
Let I = ∫ \(\sqrt{\sin 2 x}\) cos 2x dx
Putting sin 2x = t
⇒ 2 cos 2x dx = dt
⇒ cos 2x dx = \(\frac{1}{2}\) dt

Question 28.
\(\frac{\cos x}{\sqrt{1+\sin x}}\)
Answer:

Question 29.
cot x log sin x
Answer:
Let I = ∫ cot x log sin x dx
Putting log sin x = t
⇒ \(\frac{1}{\sin x}\) × cos x dx = dt
⇒ cot x dx = dt
∴ ∫ cot x (log sin x) dx = ∫t dt = \(\frac{t^{2}}{2}\) + C
= \(\frac{(\log \sin x)^{2}}{2}\) + C

Question 30.
\(\frac{\sin x}{1+\cos x}\)
Answer:
Let I = ∫ \(\frac{\sin x}{1+\cos x}\) dx
Putting 1 + cos x = t
⇒ - sin x dx = dt
⇒ sin x dx = - dt
∴ ∫ \(\frac{\sin x}{1+\cos x}\) dx = - ∫ \(\frac{1}{t}\) dt
= - log |t| + C
= - log |(1 + cos x)| + C

Question 31.
\(\frac{\sin x}{(1+\cos x)^{2}}\)
Answer:
Let I = ∫ \(\frac{\sin x}{(1+\cos x)^{2}}\) dx
Putting 1 + cos x = t
⇒ - sin x dx = dt
⇒ sin x dx = - dt

Question 32.
\(\frac{1}{(1+\cot x)}\)
Answer:

Question 33.
\(\frac{1}{1-\tan x}\)
Answer:

Question 34.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Answer:

Question 35.
\(\frac{(1+\log x)^{2}}{x}\)
Answer:

Question 36.
\(\frac{(x+1)(x+\log x)^{2}}{x}\)
Answer:

Question 37.
\(\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}\)
Answer:

Question 38.
∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx equals:
(A) 10^x - x¹⁰ + C
(B) 10^x + x¹⁰ + C
(C) (10^x - x¹⁰)^-1
(D) log (10^x + x¹⁰) + C
Answer:
Let x¹⁰ + 10^x = t
⇒ (10x⁹ + 10^x log_e 10) dx = dt
∴ ∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx = ∫\(\frac{d t}{t}\)
= log |t| + C
= log (x¹⁰ + 10^x) + C
Hence, (D) is the correct answer.

Question 39.
∫ \(\frac{d x}{\sin ^{2} x \cos ^{2} x}\) equals:
(A) tan x + cot x + C
(B) tan x - cot x + C
(C) tan x cot x + C
(D) tan x - cot 2x + C
Answer:

Hence, (B) is the correct answer.