NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.2

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

NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.2

1 से 37 तक के प्रश्नों में प्रत्येक का समाकलन ज्ञात कीजिए।
प्रश्न 1.
\(\frac{2 x}{1+x^2}\)
हल:
\(\frac{2 x}{1+x^2}\) dx
माना 1 + x² = t
तब 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

प्रश्न 2.
\(\frac{(\log x)^2}{x}\)
हल:
\(\frac{(\log x)^2}{x}\) dx
माना
log.x = t
तब
\(\frac{1}{x}\) dx = dt
∴ ∫\(\frac{(\log x)^2}{x}\) dx = t² dt
= \(\frac{t^3}{3}\) + C
= [\(\frac{(\log x)^3}{3}\) + C

प्रश्न 3.
\(\frac{1}{(x+x \log x)}\)
हल:
\(\frac{1}{(x+x \log x)}\) dx
= ∫ \(\frac{d x}{x(1+\log x)}\)
माना
1 + log x = t
तब
\(\frac{1}{x}\) dx = dt
∴ ∫\(\frac{d x}{x(1+\log x)}\) = ∫\(\frac{d t}{t}\)
= log t + C
= log |(1 + log x)| + C

प्रश्न 4.
sin x sin (cos x)
हल:
∫sin x sin (cos x) dx
माना cos x = t
तब
– sin x dx = dt
⇒ sin x dx = – dt
∴ ∫sin x sin (cos x) dx = – sint dt
= -(-cost) + C
= cos (cos x) + C

प्रश्न 5.
sin (ax + b) cos (ax + b)
हल:
sin (ax + b) cos (ax + b) dx
माना
sin (ax + b) = 1
तब acos (ax + b) dx = dt
या
cos (ax + b) dx = \(\frac{d t}{a}\)
∴ ∫sin (ax + b) cos (ax + b) dx
= ∫\(\frac{t d t}{a}\) = \(\frac{1}{a}\)∫t dt
= \(\frac{1}{a}\) \( \frac{t^2}{2}\) + C
= \(\frac{1}{2a}\) sin² (ax + b) + C

प्रश्न 6.
√ax + b
हल:
√ax + b dx = ∫(ax + b)^1/2 dx
माना
ax + b = t
तब
a dx = dt
या
dx = \(\frac{d t}{a}\)
∴ ∫(ax+b)^1/2 dx = ∫\(\frac{t^{1 / 2}}{a}\) dt
= \(\frac{1}{a}\) . \(\frac{t^{3 / 2}}{3 / 2}\) + C
= \(\frac{2}{3 a}\) (ax + b)^3/2 + C

प्रश्न 7.
x√x+2
हल:
x√x+2 dx
माना x + 2 = t
⇒ x = t – 2
तब
∴ ∫x√x+2 dx = √(t – 2)t^1/2dt
= ∫(t^1/2 – 2t^1/2) dt
= ∫t^3/2 dt – 2∫t^1/2 dt
= \(\frac{t^{5 / 2}}{\frac{5}{2}}\) – \(\frac{2 . t^{3 / 2}}{\frac{3}{2}}\) + C
= \(\frac{2}{5}\)(x + 2)

प्रश्न 8.
x√1 + 2x²
हल:
∫x√1 + 2x² dx
माना 1 + 2x²= t
तब 4x dx = dt
या
x dx = \(\frac{d t}{4}\)
∴ ∫x√1 + 2x² dx = ∫\(\frac{t^{1 / 2}}{4}\) dt
= \(\frac{1}{4}\) ∫t^1/2 dt
= \(\frac{1}{4}\) \(\frac{t^{3 / 2}}{3 / 2}\) + C
= \(\frac{2}{4 \times 3}\)(1 + 2x²)^3/2 + C
= \(\frac{1}{6}\) (1 + 2x²)^3/2 + C

प्रश्न 9.
(4x + 2) √x² + x + 1
हल:
(4x + 2) √x² + x + 1 dx
= ∫2(2x + 1) √x² + x + 1
माना x² + x + 1 = t
तब (2x + 1) dx = dt
∴ 2∫(2x + 1) √x² + x + 1 dx
= 2∫ t^1/2 dt
= 2 \(\frac{t^{3 / 2}}{3 / 2}\) + C
= \(\frac{4}{3}\)(x² + x + 1)^3/2 + C

प्रश्न 10.
\(\frac{1}{x-\sqrt{x}}\)
हल:
\(\frac{1}{x-\sqrt{x}}\) dx
= ∫\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}\) dx
माना √x – 1 = t
तब \( \frac{1}{2 \sqrt{x}}\) dx = dt
या
\(\frac{d x}{\sqrt{x}}\) = 2dt
∴∫\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}\) dx = ∫\(\frac{2 d t}{t}\)
= 2 ∫\(\frac{d t}{t}\) = 2 log |t| + C
= 2 log |√x – 1| + C

प्रश्न 11.
\(\frac{x}{\sqrt{x+4}}\)
हल:

प्रश्न 12.
(x³ – 1 )^1/3 x⁵
हल:
∫(x³ – 1 )^1/3 x⁵ dx
माना x³ – 1 = t
तथा x³ = t + 1
तब
3x² dx = dt
या
x³ dx = \(\frac{d t}{3}\)
∴ ∫(x³ – 1 )^1/3 x⁵ dx

प्रश्न 13.
\(\frac{x^2}{\left(2+3 x^3\right)^3}\)
हल:
\(\frac{x^2}{\left(2+3 x^3\right)^3}\) dx
माना
2 + 3x³ = t
तब
9x² dx = dt
या
x² dx = \(\frac{1}{9}\) dt
∴ ∫\(\frac{x^2}{\left(2+3 x^3\right)^3}\) dx
= \(\frac{1}{9}\) ∫\(\frac{d t}{t^3}\)
= \(\frac{1}{9}\)∫t^-3dt
= \(\frac{1}{9}\) \(\frac{t^{-3+1}}{(-3+1)}\) + C
= \(\frac{1}{9}\) (\(\frac{t^{-2}}{-2}\)) + C
= –\(\frac{1}{18}\)t-² + C
= – \(\frac{1}{18 t^2}\) + C
= – \(\frac{1}{18\left(2+3 x^3\right)^2}\) + C

प्रश्न 14.
\(\frac{1}{x(\log x)^m}\),x > 0, m = 1
हल:
∫\(\frac{1}{x(\log x)^m}\) dx
माना
log x = t
तब
\(\frac{1}{x}\) dx = dt
∴ ∫\(\frac{1}{x(\log x)^m}\) dx
= ∫\(\frac{d t}{t^m}\) = ∫t^-m dt
= \(\frac{t^{-m+1}}{-m+1}\) + C
= \(\frac{t^{1-m}}{1-m}\) + C
= \(\frac{(\log x)^{1-m}}{1-m}\) + C

प्रश्न 15.
\(\frac{x}{9-4 x^2}\)
हल:
∫\(\frac{x}{9-4 x^2}\) dx
माना
9 – 4x² = t
तब – 8x dx = dt
या
x dx = – \(\frac{1}{8}\) dt
∴ ∫ \(\frac{x}{9-4 x^2}\) dx = ∫-\(\frac{1}{8}\) .\(\frac{d t}{t}\)
= – \(\frac{1}{8}\)∫\(\frac{d t}{t}\)
= – \(\frac{1}{8}\)log|t| + C
= – \(\frac{1}{8}\)log|9 – 4x²| + C = –\(\frac{1}{8}\)log(9 – 4x²) + C

प्रश्न 16.
e^{2x + 3}
हल:
∫e^{2x + 3} dx
= e^2x e³ dx
= e³∫e^2x dx
माना 2x = t
तंब 2dx = dt
या dx = \(\frac{d t}{2}\)
∴ ∫e^{2x + 3} dx = e³∫ e^t \(\frac{d t}{2}\)
= \(\frac{e^3}{2}\) e^t + C
= 1/2 e^t+3 + c
= 1/2 e^{2x + 3} + C

प्रश्न 17.
\(\frac{x}{e^{x^2}}\) dx
हल:
माना x² = t
तब 2x dx = dt
या
x dx = \(\frac{1}{2}\) dt
∴ ∫\(\frac{x}{e^{x^2}}\) dx = \(\frac{1}{2}\) ∫\(\frac{d t}{e^t}\)
= \(\frac{1}{2}\) ∫e^-tdt
= \(\frac{1}{2}\) \(\frac{e^{-t}}{-1}\) + C
= – \(\frac{1}{2}\)e^-x + C
= \(-\frac{1}{2 e^{x^2}}\) + C

प्रश्न 18.
\(\frac{e^{\tan ^{-1} x}}{1+x^2}\)
हल:
\(\frac{e^{\tan ^{-1} x}}{1+x^2}\) dx
माना tan^-1 x = t
तब
\(\frac{1}{1+x^2}\) dx = dt
∴ ∫ \(\frac{e^{\tan ^{-1} x}}{1+x^2}\) dx = ∫e^t dt
= e^t + C
= e^tan-1x + C

प्रश्न 19.
\(\frac{e^{2 x}-1}{e^{2 x}+1}\)
हल:
∫\(\frac{e^{2 x}-1}{e^{2 x}+1}\) dx
= ∫ \(\frac{e^x\left(e^x-e^{-x}\right)}{e^x\left(e^x+e^{-x}\right)}\) dx
= ∫ \(\frac{e^x-e^{-x}}{e^x+e^{-x}}\) dx
(अंश तथा हर में से e^x उभयनिष्ठ लेने पर)
माना
e^x + e^-x = 1
तब
(e^x – e-^-x) dx = dt
∴ ∫\(\frac{e^{2 x}-1}{e^{2 x}+1}\) dx = ∫\(\frac{e^x-e^{-x}}{e^x+e^{-x}}\) dx
= ∫\(\frac{d t}{t}\)
= log|t| + C
= log|e^x – e-^-x| + C

प्रश्न 20.
\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
हल:
\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx
माना
e^2x + e^-2x = t
तब
(2e^2x – 2e^-2x) dx = dt
या
2(e^2x – e^-2x) dx = dt
या (e^2x – e^-2x) = \(\frac{1}{2}\) dt
∴ ∫\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx = \(\frac{1}{2}\)∫\(\frac{d t}{t}\)
= \(\frac{1}{2}\) log|t| + C
= log(e^2x + e^-2x)^1/2 + C

प्रश्न 21.
tan² (2x – 3)
हल:
tan² (2x – 3) dx
= ∫(sec² (2x – 3) – 1} dx
= ∫sec² (2x – 3) dx – ∫ dx
माना
2x – 3 = t
तब
2dx = dt
या
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

प्रश्न 22.
sec² (7 – 4x)
हल:
∫sec² (7 – 4x) dx
माना 7 – 4x = t
तब -4 dx = dt
या
dx = –\(\frac{1}{2}\) dt
∴ ∫sec² (7 – 4x) dx
= –\(\frac{1}{2}\) ∫sec² t dt
= –\(\frac{1}{2}\)tan + C
= –\(\frac{1}{2}\)tan (7 – 4x) + C

प्रश्न 23.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\)
हल:
∫\(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\) dx
माना
sin^-1 x = t
तब
\(\frac{1}{\sqrt{1-x^2}}\) dx = dt
∴ ∫\(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\) dx = ∫t dt
= \(\frac{t^2}{2}\) + C
= \(\frac{\left(\sin ^{-1} x\right)^2}{2}\) + C

प्रश्न 24.
\(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
हल:
∫\(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx
= ∫\(\frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}\) dx
= \(\frac{1}{2}\) ∫\(\frac{2 \cos x-3 \sin x}{3 \cos x+2 \sin x}\) dx
माना 3 cos x + 2 sin x = t
या (-3 sin x + 2 cos x) dx = dt
या (2 cos x – 3 sin x) dx = dt
∴ \(\frac{1}{2}\)∫\(\frac{2 \cos x-3 \sin x}{3 \cos x+2 \sin x}\) dx
= \(\frac{1}{2}\)∫\(\frac{d t}{t}\)
=\(\frac{1}{2}\) log|t| + C
= \(\frac{1}{2}\) log | 3 cos x + 2 sin x| + C

प्रश्न 25.
\(\frac{1}{\cos ^2 x(1-\tan x)^2}\)
हल:
∫\(\frac{1}{\cos ^2 x(1-\tan x)^2}\) dx
= ∫\(\frac{\sec ^2 x}{(1-\tan x)^2}\) dx
माना 1- tan x = t
तब – sec² x dx = dt
या sec² x dx = – dt
∴ ∫\(\frac{\sec ^2 x}{(1-\tan x)^2}\) dx = ∫ – \(\frac{d t}{t^2}\)
= – ∫\(\frac{d t}{t^2}\) = -∫t ^-2 dt
= \(\frac{-t^{-2+1}}{-2+1}\) + C
= –\(\frac{-t^{-1}}{-1}\) + C
= t^-1 + C
= \(\frac{1}{(1-\tan x)}\) + C

प्रश्न 26.
\(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
हल:
∫\(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
माना
√x = t
तब
\(\frac{1}{2 \sqrt{x}}\) dx = dt
या
\(\frac{1}{\sqrt{x}}\) dx = 2dt
∴ ∫\(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx = 2∫cost dt
= 2 sin t + C
= 2 sin \(\sqrt{x}\) + C

प्रश्न 27.
√sin 2x cos 2.x
हल:
∫ √sin 2x cos 2x dx
माना
sin 2x = t
तब
2 cos 2x dx = dt
या
cos 2x dx = \(\frac{1}{2}\) dt
∴ ∫√sin 2x cos 2x dx
= \(\frac{1}{2}\)∫t^1/2 dt
= \(\frac{1}{2}\) \(\frac{t^{1 / 2+1}}{\frac{1}{2}+1}\) + C
= \(\frac{1}{2}\)\(\frac{t^{3 / 2}}{3 / 2}\) + C
= \(\frac{1}{2}\) \(\frac{2}{3}\) (sin 2x)^3/2 + C
= \(\frac{1}{3}\) (sin 2x)^3/2 + C

प्रश्न 28.
\(\frac{\cos x}{\sqrt{1+\sin x}}\)
हल:
∫\(\frac{\cos x}{\sqrt{1+\sin x}}\) dx
माना 1 + sin x = t
तब cos x dx = dt
∴ ∫\(\frac{\cos x}{\sqrt{1+\sin x}}\) dx = ∫\(\frac{d t}{t^{1 / 2}}\)
= ∫t^-1/2 dt
= \(\frac{t^{-1 / 2+1}}{-\frac{1}{2}+1}\) + C
= \(\frac{t^{1 / 2}}{\frac{1}{2}}\) + C
= 2(1+ sin x)^1/2 + C

प्रश्न 29.
cot x log (sin x)
हल:
∫cot x log (sin x) dx
माना
log sin x = t
तब
\(\frac{1}{\sin x}\) x cos x dx = dt
या
cot x dx = dt
∴ ∫ cot x (log sin x) dx = ∫ t dt = \(\frac{t^2}{2}\) + C
= \(\frac{(\log \sin x)^2}{2}\) + C

प्रश्न 30.
\(\frac{\sin x}{1+\cos x}\)
हल:
∴ ∫\(\frac{\sin x}{1+\cos x}\) dx
माना 1+ cos x = t
तब – sin x dx = dt
∴ ∫\(\frac{\sin x}{1+\cos x}\) dx = -∫\(\frac{d t}{t}\)
= – log|t| + C
= – log (1 + cos x) + C
= log|\(\frac{1}{1+\cos x}\)| + C

प्रश्न 31.
\(\frac{\sin x}{1+\cos x)^2}\)
हल:
∴ ∫\(\frac{\sin x}{1+\cos x)^2}\) dx
माना 1 + cos x = t
तब – sin x dx = dt
या sin x dx =- dt
∴ ∫\(\frac{\sin x}{1+\cos x)^2}\) dx = -∫\(\frac{d t}{t}\)
= -∫t^-2 dt
= – \(\frac{t^{-2+1}}{-2+1}\) + C
= – \(\frac{t^{-1}}{-1}\) + C
= \(\frac{1}{t}\) + C
= \(\frac{1}{(1+\cos x)}\) + C

प्रश्न 32.
\(\frac{1}{1+\cot x}\)
हल:

प्रश्न 33.
\(\frac{1}{1-\tan x}\)
हल:

माना cosx – sinx = t
तब (-sinx – cosx) dx dt
या -(sinx + cosx ) dx = dt
या (sinx + cosx ) dx = -dt
∴ \(\frac{1}{2}\)∫dx + \(\frac{1}{2}\)∫\(\frac{\cos x+\sin x}{\cos x-\sin x}\) dx
= \(\frac{1}{2}\)∫dx + \(\frac{1}{2}\)∫\(\frac{(-d t)}{t}\)
= \(\frac{1}{2}\)∫dx – \(\frac{1}{2}\)∫\(\frac{(d t)}{t}\)
= \(\frac{1}{2}\)x – \(\frac{1}{2}\) log|t| + C
= \(\frac{1}{2}\)x – \(\frac{1}{2}\)log|cosx – sinx| + C
= \(\frac{x}{2}\) – \(\frac{1}{2}\)log|cosx – sinx| + C

प्रश्न 34.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
हल:

प्रश्न 35.
\(\frac{(1+\log x)^2}{x}\)
हल:
∫\(\frac{(1+\log x)^2}{x}\) dx
माना 1 + log x = t
तब \(\frac{1}{x}\) dx = dt
∴∫ \(\frac{(1+\log x)^2}{x}\) dx
= ∫t² dt = \(\frac{t^3}{3}\) + C
= \(\frac{(1+\log x)^3}{3}\) + C
= \(\frac{1}{3}\) (1 + logx)³ + C

प्रश्न 36.
\(\frac{(x+1)(x+\log x)^2}{x}\)
हल:
∴∫\(\frac{(x+1)(x+\log x)^2}{x}\) dx
= ∫\(\left(\frac{x+1}{x}\right)\)(x + logx)² dx
माना x + logx = t
तब \(\left(\frac{x+1}{x}\right)\) dx = dt
या \(\left(\frac{x+1}{x}\right)\) dx = dt
∴∫\(\left(\frac{x+1}{x}\right)\)(x + logx)² dx = ∫t² dt
= \(\frac{t^3}{3}\) + C
= \(\frac{(x+\log x)^3}{3}\) + C
= \(\frac{1}{3}\) (x + log x)³ + C

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

प्रश्न 38.
∫ \(\frac{10 x^9+10^x \log _e 10}{x^{10}+10^x}\) dx
(A) 10^x – x¹⁰ + C
(B) 10^x + x¹⁰ + C
(C) (10^x + x¹⁰) + C
(D) log(10^x + x¹⁰) + C
हल:
माना 10^x + x¹⁰ = t
तब (10x⁹ + 10^x loge 10) dx = dt
= log|t| + C
= log(x¹⁰ + 10^x) + C
∴∫\(\frac{10 x^9+10^x \log _e 10}{x^{10}+10^x}\) dx = ∫\(\frac{d t}{t}\)
= log|t| + C = log(x¹⁰ + 10^x) + C
अत: विकल्प (D) सही है।

प्रश्न 39.
∫\(\frac{d x}{\sin ^2 x \cos ^2 x}\) dx
(A) tanx + cot x + C
(B) tanx – cot x + C
(C) tanxcotx + C
(D) tanx – cot2x + C
हल:
∫\(\frac{d x}{\sin ^2 x \cos ^2 x}\) dx
= ∫\(\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}\) dx
(∴ sinx² + cosx² ) + C
= ∫\(\frac{\sin ^2 x}{\sin ^2 x \cos ^2 x}\)dx + ∫\(\frac{\cos ^2 x}{\sin ^2 x \cos ^2 x}\)dx
= ∫\(\frac{1}{\cos ^2 x}\) dx + ∫\(\frac{1}{\sin ^2 x}\) dx
= ∫sec²x dx + ∫cosec²x dx
= tanx – cotx + c
अतः विकल्प (B)सही है।