These NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.3 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.3
1 से 22 तक के प्रश्नों में प्रत्येक फलन का समाकलन ज्ञात कीजिए।
प्रश्न 1.
sin2 (2x + 5)
हल:
∫sin2 (2x + 5) dx
= ∫ \(\frac{1-\cos 2(2 x+5)}{2}\) dx
[∵ cos 2x = 1 – 2 sin2x]
= \(\frac{1}{2}\)∫dx – \(\frac{1}{2}\)∫cos(4x + 10)dx
= \(\frac{x}{2}\) – \(\frac{1}{2}\) \(\frac{\sin (4 x+10)}{4}\) + C
= \(\frac{x}{2}\) – \(\frac{\sin (4 x+10)}{8}\) + C
प्रश्न 2.
sin 3x cos 4x
हल:
∫sin 3x cos 4x dx
= \(\frac{1}{2}\) ∫2 sin3xcos4xdx
= \(\frac{1}{2}\) ∫[sin(3x + 4x) + sin(3x – 4x)]dx
[∵2 sin A cos B = sin (A + B) + sin (A – B)]
= \(\frac{1}{2}\) ∫ [sin 7x + sin(-x)) dx
= \(\frac{1}{2}\) ∫sin 7x dx – = \(\frac{1}{2}\) ∫ sinx dx
[∵sin(x) = sin x]
= \(\frac{1}{2}\) \(\frac{(-\cos 7 x)}{7}\) – \(\frac{1}{2}\)(-cosx) + C
= –\(\frac{1}{2}\)Cos7x + \(\frac{1}{2}\)cosx + C
प्रश्न 3.
cos 2x cos 4x cos 6x
हल:
∫cos 2x cos 4x cos 6x dx
= \(\frac{1}{2}\) ∫(2cos 2x cos 4x) cos 6x dx
= \(\frac{1}{2}\) ∫cos (2x + 4x) + cos (2x – 4x)] cos 6x dx
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= \(\frac{1}{2}\) ∫cos 6x + cos (-2x)] cos 6x dx
= \(\frac{1}{2}\) ∫[cos 6x + cos 2x] cos 6x dx
[∵cos (θ) = cosθ]
= \(\frac{1}{2}\) ∫cos2 6x + \(\frac{1}{2}\) ∫cos 2x cos 6x dx
= \(\frac{1}{4}\) ∫cos2 6x dx + \(\frac{1}{4}\) ∫2 cos 2x cos 6x dx
= \(\frac{1}{4}\) ∫ (1 + cos 12x) dx + \(\frac{1}{4}\) ∫[cos (2x + 6x) + cos (2x – 6x)] dx
= \(\frac{1}{4}\) ∫ (1 + cos 12x) dx + [cos 8x + cos (-4x)] dx
= \(\frac{1}{4}\) ∫ (1 + cos 12x) dx + (cos 8x + cos 4x) dx
= [cos (-θ) = cosθ]
= \(\frac{1}{4}\) ∫dx + = \(\frac{1}{4}\) ∫cos12xdx + = \(\frac{1}{4}\) ∫cos8xdx + \(\frac{1}{4}\) ∫cos4x dx
= \(\frac{1}{4}\)x + \(\frac{1}{4}\)
प्रश्न 4.
sin3 (2x + 1)
हल:
∫sin3 (2x + 1) dx
= ∫\(\frac{3 \sin (2 x+1)-\sin 3(2 x+1)}{4}\) dx
[∵sin 3x = 3 sin x – 4 sin3x]
= \(\frac{3}{4}\)∫sin(2x + 1) dx – \(\frac{1}{4}\)∫sin(6x + 3) dx
= \(\frac{3}{4}\)\(\frac{[-\cos (2 x+1)]}{2}\) – \(\frac{1}{4}\)\(\left(\frac{-\cos (6 x+3)}{6}\right)\) + C
= \(\frac{3}{8}\)cos (2x + 1) + \(\frac{1}{24}\)cos(6x + 3) + C
= –\(\frac{3}{8}\)cos3 (2x + 1) + \(\frac{1}{24}\)cos 3(2x + 1) + C
= –\(\frac{3}{8}\)cos (2x + 1) + \(\frac{1}{24}\)[4 cos3(2x + 1) – 3cos(2x + 1)] + C
[∵ cos3x = 4 cos3x – 3cosx]
= –\(\frac{3}{8}\)cos (2x + 1) + \(\frac{1}{6}\)cos3(2x + 1) \(\frac{1}{8}\)cos (2x + 1) + C
= –\(\frac{1}{2}\)cos (2x + 1) + \(\frac{1}{6}\)cos (2x + 1) + C
प्रश्न 5.
sin3 x cos3 x
हल:
∫sin3 x cos3 x dx
= ∫sin x sin2x cos3x dx
= ∫sin x (1 – cos2 x) cos3 x dx
माना cos x = t
तब – sin x dx = dt
या
sin x dx = – dt
∴ ∫sin x (1 – cos2 x)cos3 x dx
= -∫(1 – t2)t3dt
= -∫(t3 – t5)dt
= ∫t5dt – ∫t3dt
= \(\frac{t^6}{6}\) – \(\frac{t^4}{4}\) + C
= \(\frac{1}{6}\)cos6x – \(\frac{1}{4}\)cos4x + C
प्रश्न 6.
sin x sin 2x sin 3x
हल:
∫sin x sin 2x sin 3x dx
= \(\frac{1}{2}\) ∫2sin x sin 2x sin 3x dx
= \(\frac{1}{2}\) ∫(2 sin x sin 2x) sin 3x dx
= \(\frac{1}{2}\) ∫[cos(x – 2x) – cos(x + 2x)] sin 3x dx
[ 2 sin A sin B = cos (A – B) – cos (A + B)
= \(\frac{1}{2}\) ∫[cos(-x) – cos3x] sin3x dx
= \(\frac{1}{2}\) ∫[(cosx – cos 3x) sin 3x dx
[∵ cos (-θ) = cos θ)]
= \(\frac{1}{2}\) ∫cos xsin3x dx – \(\frac{1}{2}\) ∫sin 3x cos x dx
= \(\frac{1}{4}\) ∫2cos3xsin3x dx – \(\frac{1}{4}\) ∫cos 3x sin 3x dx
= \(\frac{1}{4}\) ∫(sin (3x – x) + sin (3x – x)] dx – \(\frac{1}{4}\) ∫sin 6x dx
[∵ 2 sin A cos B = sin (A + B) + sin (A – B)]
= \(\frac{1}{4}\) ∫(sin 4x + sin 2x) dx – \(\frac{1}{4}\) ∫sin 6x dx
= \(\frac{1}{4}\) ∫sin 4x dx + \(\frac{1}{2}\) ∫sin 2x dx – \(\frac{1}{4}\) ∫sin6xdx
= \(\frac{1}{4}\)\(\frac{(-\cos 4 x)}{4}\) + \(\frac{1}{4}\)\(\frac{(-\cos 2 x)}{2}\) – \(\frac{1}{4}\)\(\frac{(-\cos 6 x)}{6}\) + C
= –\(\frac{1}{16}\)cos 4x – \(\frac{1}{8}\)cos 2x + \(\frac{1}{24}\)cos 6x + C
= –\(\frac{1}{4}\) \(\frac{1}{6}\)cos6x – \(\frac{1}{4}\)cos4x – \(\frac{1}{2}\)cos2x + C
प्रश्न 7.
sin 4x sin &x
हल:
∫sin 4x sin 8x dx
= \(\frac{1}{2}\) ∫2sin 4x sin 8x dx
\(\frac{1}{2}\) ∫ [cos (4x – 8x) – cos (4x + 8x)] dx
[∵ 2 sin A sin B = cos (AB) – cos (A + B)]
= \(\frac{1}{2}\) ∫[cos (-4x) – cos 12x] dx
= \(\frac{1}{2}\) ∫[cos 4x – cos 12x] dx
[∵ cos (-θ) = cos θ)]
= \(\frac{1}{2}\) ∫cos 4x dx – \(\frac{1}{2}\) ∫ cos 12x dx
= \(\frac{1}{2}\) \(\frac{\sin 4 x}{4}\) – \(\frac{1}{2}\) \(\frac{\sin 12x}{12}\) + C
= \(\frac{1}{8}\)sin 4x – \(\frac{1}{24}\)sin12x + C
= \(\frac{1}{2}\)[\(\frac{1}{4}\)sin4x – \(\frac{1}{12}\)sin12x] + C
प्रश्न 8.
\(\frac{1-\cos x}{1+\cos x}\)
हल:
∫\(\frac{1-\cos x}{1+\cos x}\) dx
= ∫\(\frac{2 \sin ^2 x / 2}{2 \cos ^2 x / 2}\)dx
= ∫tan2 \(\frac{x}{2}\) dx
= ∫(sec2 \(\frac{x}{2}\) – 1) dx
= ∫sec2\(\frac{x}{2}\) dx – ∫dx
= 2tan\(\frac{x}{2}\) – x + C
प्रश्न 9.
\(\frac{\cos x}{1+\cos x}\)
हल:
∫\(\frac{\cos x}{1+\cos x}\) dx
= ∫\(\frac{1+\cos x-1}{1+\cos x}\)dx
( अंश में 1 जोड़ने तथा घटाने पर )
= ∫\(\frac{1+\cos x}{1+\cos x}\)dx – ∫\(\frac{1}{1+cos x}\)dx
= ∫dx – ∫\(\frac{1}{2 \cos ^2 \frac{x}{2}}\)dx
= ∫dx – \(\frac{1}{2}\)∫sec2 \(\frac{x}{2}\)dx
= x – \(\frac{1}{2}\)∫sec2\(\frac{x}{2}\)dx
= x – \(\frac{1}{2}\) × 2tan\(\frac{x}{2}\) + C
= x – tan\(\frac{x}{2}\) + C
प्रश्न 10.
sin4 dx
हल:
∫sin4 dx
= ∫(sin2 x)2 dx
= ∫\(\left(\frac{1-\cos 2 x}{2}\right)^2\)dx
= \(\frac{1}{4}\)∫[1 + cos22x – 2cos2x]dx
= \(\frac{1}{4}\)[∫dx + ∫cos22x dx – 2∫co 2x dx]
= \(\frac{1}{4}\)[∫dx + ∫\(\frac{1+\cos 4 x}{2}\) dx – 2∫cos2x dx
= \(\frac{1}{4}\)∫dx + \(\frac{1}{8}\)∫dx + \(\frac{1}{8}\)∫cos 4x dx – \(\frac{1}{2}\)∫cos 2x dx
= \(\frac{3}{8}\)∫dx + \(\frac{1}{8}\)∫cos 4x dx – \(\frac{1}{2}\)∫cos 2x dx
= \(\frac{3}{8}\)x + \(\frac{1}{8}\) \(\frac{\sin 4 x}{4}\) – \(\frac{1}{2}\) \(\frac{\sin 2x}{2}\) + C
= \(\frac{3}{8}\)x + \(\frac{1}{32}\)sin 4x – \(\frac{1}{4}\)sin 2x + C
= \(\frac{3x}{8}\) – \(\frac{1}{4}\)sin 2x + \(\frac{1}{32}\) sin 4x + C
प्रश्न 11.
cos4 2x
हल:
∫cos4 2x dx
= ∫(cos2 2x)2 dx
= ∫\(\left(\frac{1+\cos 4 x}{2}\right)^2\) dx
= \(\frac{1}{4}\)∫(1 + cos2 4x + 2cos 4x) dx
= \(\frac{1}{4}\)∫dx + \(\frac{1}{4}\)∫ cos2 4x dx + \(\frac{1}{4}\)∫2 cos 4x dx
= \(\frac{1}{4}\)∫dx + \(\frac{1}{4}\)∫\(\frac{1+\cos 8 x}{2}\) dx + \(\frac{1}{2}\)∫cos 4x dx
= \(\frac{1}{4}\)∫dx + \(\frac{1}{4}\)∫dx + \(\frac{1}{8}\)∫cos 8x dx + \(\frac{1}{2}\)∫cos 4x dx
= \(\frac{3}{8}\)∫dx + \(\frac{1}{8}\)∫cos 8x dx + \(\frac{1}{2}\)∫cos 4x dx
= \(\frac{3}{8}\)x + \(\frac{1}{8}\) \(\frac{\sin 8 x}{8}\) + \(\frac{1}{2}\)\(\frac{\sin 8 x}{8}\) + C
= \(\frac{3}{8}\)x + \(\frac{1}{64}\)sin 8x + \(\frac{1}{8}\)sin 4x + C
= \(\frac{3}{8}\)x + \(\frac{1}{8}\)sin 4x + \(\frac{1}{64}\)sin 8x + C
प्रश्न 12.
\(\frac{\sin ^2 x}{1+\cos x}\)
हल:
∫\(\frac{\sin ^2 x}{1+\cos x}\) dx
= ∫\(\frac{1-\cos ^2 x}{1+\cos x}\) dx
= ∫\(\frac{(1-\cos x)(1+\cos x)}{(1+\cos x)}\)
= ∫(1 + cos x) dx
= ∫dx – ∫cos x dx
= x – sin x + C
प्रश्न 13.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
हल:
∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
= ∫\(\frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha}\)
= ∫\(\frac{\left(2 \cos ^2 x-1-2 \cos ^2 \alpha+1\right)}{\cos x-\cos \alpha}\) dx
= 2∫\(\frac{\cos ^2 x-\cos ^2 \alpha}{\cos x-\cos \alpha}\) dx
= 2∫\(\frac{(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{(\cos x-\cos \alpha)}\) dx
= 2∫(cos x + cosα) dx
= 2∫cos x dx + 2∫cosα dx
= 2 sin x + 2x cosα + C
= 2(sin x + x cosα) + C
प्रश्न 14.
\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
हल:
∫\(\frac{\cos x-\sin x}{1+\sin 2 x}\) dx
माना cos x + sin x = t
तब (- sin x + cos x) dx = dt
या (cos.x – sin x) dx = dt
प्रश्न 15.
tan3 2x sec 2x
हल:
∫tan3 2x sec 2x dx
= ∫tan2 2x.tan 2x sec 2x dx
= ∫(sec2 2x – 1) tan 2x sec 2x dx
माना
sec 2x = t
तब 2 sec 2x tan 2x dx = dt
या
sec 2x tan 2x dx = \(\frac{1}{2}\) dt
∴ ∫(sec2 2x – 1) tan 2x sec 2x dx
= \(\frac{1}{2}\)∫(t2 – 1)dt
= \(\frac{1}{2}\)∫(t2 – \(\frac{1}{2}\)∫dt
= \(\frac{1}{2}\)\(\frac{t^3}{3}\) – \(\frac{1}{2}\)t + C
= \(\frac{1}{6}\)sec3 2x – \(\frac{1}{2}\)sec 2x+C
प्रश्न 16.
tan4 x
हल:
∫tan4 x dx
= ∫tan2 x.tan2 x dx
= ∫tan2 x (sec2 x – 1) dx
= ∫tan2 x sec2 x dx – ∫tan2 x dx
= ∫tan2 x sec2 x dx – ∫(sec2 x – 1) dx
= ∫tan2 x sec2 x dx – ∫sec2 x dx + ∫dx ……………..(1)
माना tan x = t
तब
sec2 x dx = dt
∴ ∫tan2 x sec2 x dx = ∫t2 dt
= \(\frac{t^3}{3}\) = \(\frac{\tan ^3 x}{3}\)
तब समीकरण (1) से,
∴ ∫tan2 x sec2 x dx – ∫sec2 x dx + ∫dx
= \(\frac{\tan ^3 x}{3}\) – tan x + x + C
= \(\frac{1}{3}\) tan3 x – tan x + x + C
प्रश्न 17.
\(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\)
हल:
∫\(\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
= ∫\(\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}\) dx + ∫\(\frac{\cos ^3 x}{\sin ^2 x \cos ^2 x} \) dx
= ∫\(\frac{\sin x}{\cos ^2 x}\) dx + ∫\(\frac{\cos x}{\sin ^2 x}\) dx
= ∫sec x tan x dx + ∫cosec x cot x dx
= sec x – cosec x + C
प्रश्न 18.
\(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\)
हल:
∫ \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) dx
= ∫\(\frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x}\) dx
= ∫\(\frac{1}{\cos ^2 x}\) dx
=∫sec2 x dx
= tan x + C
प्रश्न 19.
\(\frac{1}{\sin x \cos ^3 x}\)
हल:
∫\(\frac{1}{\sin x \cos ^3 x}\) dx
= ∫\(\frac{\cos ^2 x+\sin ^2 x}{\sin x \cos ^3 x}\) dx
= ∫\(\frac{\cos ^2 x}{\sin x \cos ^3 x}\)dx + ∫\(\frac{\sin ^2 x}{\sin x \cos ^3 x}\)dx
= ∫\(\frac{\cos x}{\sin x \cos ^2 x}\) dx + ∫\(\frac{\sin x}{\cos ^3 x}\) dx
= ∫\(\frac{\sec ^2 x}{\tan x}\) dx + ∫sec2 x tan x dx
माना tan x = t
तब sec2 x dx = dt
∴ ∫\(\frac{\sec ^2 x}{\tan x}\) dx + [sec2 x tan x dx
= ∫\(\frac{1}{t}\) dt + ∫t dt
= log |t| + \(\frac{t^2}{2}\) + C
= log|tan x| + \(\frac{1}{2}\)tan2x + C
प्रश्न 20.
\(\frac{\cos 2 x}{(\cos x+\sin x)^2}\)
हल:
= ∫\(\frac{\cos 2 x}{(\cos x+\sin x)^2}\) dx
= ∫\(\frac{\left(\cos ^2 x-\sin ^2 x\right)}{(\cos x+\sin x)^2}\) dx
= ∫\(\frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^2}\) dx
= ∫\(\frac{\cos x-\sin x}{\cos x+\sin x}\) dx
log | tan x+tan2x+C
माना cos x + sin x = t
(sin x + cos x) dx = dt
या
(cos x – sin x) dx = dt
∴∫\(\frac{\cos x-\sin x}{\cos x+\sin x}\) dx
= ∫\(\frac{d t}{t}\) = log|t| + C
= log | cos x + sin x| + C
प्रश्न 21.
sin -1 (cos x)
हल:
= ∫sin -1 (cos x) dx
= ∫sin-1 [sin (\(\frac{\pi}{2}\) – x)] dx
= [(\(\frac{\pi}{2}\) – x) dx = \(\frac{\pi}{2}\)∫dx – ∫x dx
= \(\frac{\pi}{2}\)x – \(\frac{x^2}{2}\) + C
= \(\frac{\pi}{2}\) – \(\frac{x^2}{2}\) + C
प्रश्न 22.
\(\frac{1}{\cos (x-a) \cos (x-b)}\)
हल:
प्रश्न 23 एवं 24 में सही उत्तर का चयन कीजिए-
प्रश्न 23.
\(\frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cdot \cos ^2 x}\) dx बराबर है
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) tan x + cot x + C
(D) tan x + secx+C
हल:
∫\(\frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cdot \cos ^2 x}\) dx
= ∫\(\frac{\sin ^2 x}{\sin ^2 x \cos ^2 x}\) dx – ∫\(\frac{\cos ^2 x}{\sin ^2 x \cos ^2 x}\) dx
= ∫\(\frac{1}{\cos ^2 x}\) dx – ∫\(\frac{1}{\sin ^2 x}\) dx
= ∫sec2 x dx – ∫cosec2 x dx
tan x + cot x + C
अत: विकल्प (A) सही है।
प्रश्न 24.
∫\(\frac{e^x(1+x)}{\cos ^2\left(x e^x\right)}\) dx बराबर है:
(A) -cot (exx ) + C
(B) tan (xex ) + C
(C) tan (xex ) + C
(D) cot (ex ) + C
ex (1 + x)
हल:
∫\(\frac{e^x(1+x)}{\cos ^2\left(x e^x\right)}\) dx
माना xex = t
तब (xex + ex) dx = dt
या
ex (1 + x) dx = dt
∴ ∫\(\frac{e^x(1+x)}{\cos ^2\left(x e^x\right)}\) dx = ∫\(\frac{d t}{\cos ^2 t}\)
= ∫\(\)sec2 t dt
= tan t + C
= tan (xex) + C
अतः विकल्प (B) सही है।