Question

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

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

Answer 1

Answer: (C)

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}si{n}^{4}x\left(1+log\left(\frac{2+sinx}{2-sinx}\right)\right)dx..........\left(i\right)$
Use proerties $\underset{0}{\overset{b}{\int }}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$
$=\underset{-\pi /2}{\overset{\pi /2}{\int }}si{n}^{4}x\left(1+log\left(\frac{2-sinx}{2+sinx}\right)\right)dx............\left(ii\right)$
by (i) + (ii)
$\Rightarrow 2I={\int }_{-\pi /2}^{\pi /2}2si{n}^{4}xdx$
$\Rightarrow I=2{\int }_{-\pi /2}^{\pi /2}si{n}^{4}xdx=$
$=x.\frac{3.1}{4.2}.\frac{\pi }{2}=\frac{3\pi }{8}$