ISI2016-DCG-46

Question

Let $I=\int(\sin\:x-\cos\:x)(\sin\:x+\cos\:x)^{3}dx$ and $K$ be a constant of integration. Then the value of $I$ is

• A. $(\sin\:x+\cos\:x)^{4}+K$
• B. $(\sin\:x+\cos\:x)^{2}+K$
• C. $-\frac{1}{4}(\sin\:x+\cos\:x)^{4}+K$
• D. None of these

Comments

Option C is correct

Answer 1

Answer :  C

 

$I = \int \left ( \sin x - \cos x \right )\left (\sin x + \cos x \right )^{3}dx$

$I = \int \left ( \sin x - \cos x \right )\left (\sin x + \cos x \right )\left (\sin x + \cos x \right )^{2}dx$

$I = \int \left (\sin ^{2}x - \cos^{2} x \right )\left (\sin x + \cos x \right )^{2}dx$

$I = - \int \left (\cos^{2} x - \sin ^{2}x \right )\left (\sin^{2} x + \cos^{2} x + 2 \sin x \cos x \right )dx$

$I = - \int \left (\cos 2x \right )\left (1 + \sin 2x \right )dx$

 

$Let \: 1 + \sin 2x = t$

$Then \: \cos 2x \, dx \: = \: \frac{dt}{2}$

 

$I = -\int t\: \frac{dt}{2}$

$I = \frac{-\:t^{2}}{4} + K$

$I = \frac{-\: \left ( 1\:+\:\sin 2x \right )^{2}}{4} + K$

$I = \frac{-\: \left ( \sin^2x \:+ \:\cos^2x \:+\: 2\sin x \cos x \right )^{2}}{4} + K$

$I = \frac{-\: \left ( \sin x \:+\: \cos x \right)^{4}}{4} + K$