Question

The value of the integral
$\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{4}x(1+\log \left(\frac{2+\sin x}{2-\sin x}\right))dx$ is :

• A. $0$
• B. $\frac{3}{4}$
• C. $\frac{3}{8}\pi$
• D. $\frac{3}{16}\pi$

Answer 1

The correct option is C $\frac{3}{8}\pi$

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{4}x(1+\log \left(\frac{2+\sin x}{2-\sin x}\right))dx\cdots \left(1\right)\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^4(-x)(1+\log \left(\frac{2+\sin (-x)}{2-\sin (-x)}\right))dx\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{4}x(1+\log \left(\frac{2-\sin x}{2+\sin x}\right))dx\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{4}x(1-\log \left(\frac{2+\sin x}{2-\sin x}\right))dx\cdots \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right),$
$\Rightarrow 2I=\underset{-\pi /2}{\overset{\pi /2}{\int }}2{\sin }^{4}xdx\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{4}xdx\Rightarrow I=2\underset{0}{\overset{\pi /2}{\int }}{\sin }^{4}xdx\cdots \left(3\right)$
Putting $x\to \frac{\pi }{2}-x$
$\Rightarrow I=2\underset{0}{\overset{\pi /2}{\int }}{\cos }^{4}xdx\cdots \left(4\right)$
Adding (3) and (4),
$\Rightarrow 2I=2\underset{0}{\overset{\pi /2}{\int }}{\sin }^{4}x+{\cos }^{4}xdx\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}xdx\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(1{)}^2-\frac{{\sin }^{2}4x}{2}dx\Rightarrow I=\frac{\pi }{2}-4\underset{0}{\overset{\pi /8}{\int }}\frac{{\sin }^{2}4x}{2}dx$
Putting $x\to \frac{\pi }{8}-x$
$\Rightarrow I=\frac{\pi }{2}-4\underset{0}{\overset{\pi /8}{\int }}\frac{{\cos }^{2}4x}{2}dx\Rightarrow 2I=\pi -4\underset{0}{\overset{\pi /8}{\int }}\frac{1}{2}dx\Rightarrow I=\frac{3\pi }{8}$