NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.2

Question 1.
$$\frac { 2x }{ 1+{ x }^{ 2 } }$$
Solution:
Let 1 + x² = t
⇒ 2xdx = dt
∴ ∫$$\frac { 2x }{ 1+{ x }^{ 2 } }$$ dx
= ∫$$\frac { dt }{ t }$$
= log t + C
= log(1 + x²) + C

Question 2.
$$\frac { { \left( logx \right) }^{ 2 } }{ x }$$
Solution:
Let logx = t
⇒ $$\frac { 1 }{ x }dx=dt$$
$$\int \frac{(\log x)^{2}}{x} d x=\int t^{2} d t=\frac{t^{3}}{3}+\mathrm{C}$$
= $$\frac{1}{3}(\log x)^{3}+\mathrm{C}$$

Question 3.
$$\frac { 1 }{ x+xlogx }$$
Solution:
Put 1 + logx = t
∴ $$\frac { 1 }{ x }$$dx = dt
$$\int { \frac { 1 }{ x(1+logx) } dx } =\int { \frac { 1 }{ t } dt } =log|t| + C$$
= log|1+logx| + C

Question 4.
sinx sin(cosx)
Solution:
∫sinx sin(cosx)dx = ∫sin t dt
= – (-cos t) + C
= – cost + C
= cos(cos x) + C

Question 5.
sin(ax+b) cos(ax+b)
Solution:

Another method:

Question 6.
$$\sqrt { ax+b }$$
Solution:

Question 7.
x$$\sqrt { x+2 }$$
Solution:

Question 8.
x$$\sqrt { 1+{ 2x }^{ 2 } }$$
Solution:

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

Question 10.
$$\frac { 1 }{ x-\sqrt { x } }$$
Solution:

Question 11.
$$\frac { x }{ \sqrt { x+4 } }$$ , x > 0
Solution:

Question 12.
$${ { (x }^{ 3 }-1) }^{ \frac { 1 }{ 3 } }.{ x }^{ 5 }$$
Solution:

Question 13.
$$\frac { { x }^{ 2 } }{ { { (2+3x }^{ 3 }) }^{ 3 } }$$
Solution:

Question 14.
$$\frac { 1 }{ x(logx)^{ m } }$$, x > 0
Solution:
Put log x = t, $$\frac { dt }{ dx }$$ = $$\frac { 1 }{ x }$$ ⇒ dt = $$\frac { 1 }{ x }$$dx
$$\int \frac{1}{x(\log x)^{m}} d x=\int \frac{d t}{t^{m}}=\int t^{-m} d t=\frac{t^{-m+1}}{-m+1}+\mathrm{C}$$
$$=\frac { { (logx) }^{ 1-m } }{ 1-m }$$ + C

Question 15.
$$\frac { x }{ 9-4{ x }^{ 2 } }$$
Solution:

Question 16.
$${ e }^{ 2x+3 }$$
Solution:
put 2x + 3 = t
so that 2dx = dt
$$\int { { e }^{ 2x+3 } } dx\quad =\frac { 1 }{ 2 } \int { { e }^{ t }dt } \quad =\frac { 1 }{ 2 } { e }^{ t }+c\quad =\frac { 1 }{ 2 } { e }^{ 2x+3 }$$ + C

Question 17.
$$\frac { x }{ { e }^{ { x }^{ 2 } } }$$
Solution:

Question 18.
$$\frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } }$$
Solution:
let $$\quad { tan }^{ -1 }x=t\Rightarrow \frac { 1 }{ 1+{ x }^{ 2 } } dx=dt$$
∴ $$\frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } }$$dx
= ∫et dt
= et+C
= etan-1 + C

Question 19.
$$\frac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 }$$
Solution:

Another method

Question 20.
$$\frac { { e }^{ 2x }-{ e }^{ 2x } }{ { e }^{ 2x }+{ e }^{ -2x } }$$
Solution:

Question 21.
tan²(2x – 3)
Solution:
∫tan²(2x – 3)dx = ∫[sec²(2x – 3) – 1]dx = I
put 2x – 3 = t
so that 2dx = dt
I = $$\frac { 1 }{ 2 }$$ ∫sec²t dt – x + C
= $$\frac { 1 }{ 2 }$$t – x + C
= $$\frac { 1 }{ 2 }$$tan(2x -3 ) – x + C

Question 22.
sec²(7 – 4x)
Solution:
∫sec²(7 – 4x)dx
= $$\frac { tan(7-4x) }{ -4 }$$ + C

Question 23.
$$\frac { { sin }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } }$$
Solution:
Let $$\quad { sin }^{ -1 }x=t\quad\Rightarrow\frac { 1dx }{ \sqrt { 1-{ x }^{ 2 } } }$$ = dt
$$\int { \frac { { sin }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } dx } =\int { t\quad dt } =\frac { 1 }{ 2 } { t }^{ 2 }+c=\frac { 1 }{ 2 } { { (sin }^{ -1 }x) }^{ 2 }+c$$

Question 24.
$$\frac { 2cosx-3sinx }{ 6cosx+4sinx }$$
Solution:

Question 25.
$$\frac { 1 }{ { cos }^{ 2 }x{ (1-tanx) }^{ 2 } }$$
Solution:

Question 26.
$$\frac { cos\sqrt { x } }{ \sqrt { x } }$$
Solution:

Question 27.
$$\sqrt { sin2x } cos2x$$
Solution:

Question 28.
$$\frac { cosx }{ \sqrt { 1+sinx } }$$
Solution:

Question 29.
cotx log sinx
Solution:

Question 30.
$$\frac { sinx }{ 1+cosx }$$
Solution:
put 1 + cosx = t
⇒ – sinx dx = dt
∫$$\frac { sinx }{ 1+cosx }$$dx
= – $$\frac { dt }{ t }$$
= – log|t| + c
= – log|1 + cosx| + C
= $$\log \left|\frac{1}{1+\cos x}\right|+\mathrm{C}$$

Question 31.
$$\frac { sinx }{ { (1+cosx) }^{ 2 } }$$
Solution:
put 1 + cosx = t
so that – sinx dx = dt
∫$$\frac { sinx }{ { (1+cosx) }^{ 2 } }$$ = – $$\int \frac{d t}{t^{2}}=\frac{1}{t}+C$$
= $$\frac { 1 }{ t }$$ + C
= $$\frac { 1 }{ 1 + cosx }$$ + C

Question 32.
$$\frac { 1 }{ 1 + cotx }$$
Solution:

Question 33.
$$\frac { 1 }{ 1-tanx }$$
Solution:

Question 34.
$$\frac { \sqrt { tanx } }{ sinxcosx }$$
Solution:

Question 35.
$$\frac { { (1+logx) }^{ 2 } }{ x }$$
Solution:

Question 36.
$$\frac { (x+1){ (x+logx) }^{ 2 } }{ x }$$
Solution:

Question 37.
$$\frac { { x }^{ 3 }sin({ tan }^{ -1 }{ x }^{ 4 }) }{ 1+{ x }^{ 8 } } dx$$
Solution:

Question 38.
$$\int { \frac { { 10x }^{ 9 }+{ 10 }^{ x }log{ e }^{ 10 } }{ { x }^{ 10 }+{ 10 }^{ x } } dx }$$ equals
(a) 10x – x10 + C
(b) 10x + x10 + C
(c) (10x – x10) + C
(d) log (10x + x10) + C
Solution:
(d) log (10x + x10) + C

Question 39.
$$\int { \frac { dx }{ { sin }^{ 2 }x{ \quad cos }^{ 2 }x } = }$$
(a) tan x + cot x + c
(b) tan x – cot x + c
(c) tan x cot x + c
(d) tan x – cot 2 x + c
Solution:
(b) tan x – cot x + c

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