Question

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^n\theta }{{\sin }^n\theta +{\cos }^n\theta }d\theta$ is equal to

• A. $1$
• B. $0$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (D)

Let $I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^n\theta }{{\sin }^n\theta +{\cos }^n\theta }d\theta$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^n\left(\frac{\pi }{2}-\theta \right)}{{\sin }^n\left(\frac{\pi }{2}-\theta \right)+{\cos }^n\left(\frac{\pi }{2}-\theta \right)}d\theta$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\cos }^n\theta }{{\cos }^n\theta +{\sin }^n\theta }d\theta \dots (ii)$
On adding Eqs. (i) and (ii), we get
$2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^n\theta +{\cos }^n\theta }{{\cos }^n\theta +{\sin }^n\theta }d\theta =\underset{0}{\overset{\pi /2}{\int }}d\theta =\frac{\pi }{2}$
$\Rightarrow I=\frac{\pi }{4}$