Question

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

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

Answer 1

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

Let $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{4}x(1+\log \left(\frac{2+\sin x}{2-\sin x}\right))dx$ ------$i$

Now we know that,${\int }_a^{b}f\left(x\right).dx={\int }_a^{b}f(a+b-x).dx$

$\Rightarrow I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left({\sin }^4(-x)\right).(1+\log (\frac{2+\sin (-x)}{2-\sin (-x)}\left)\right).dx$

$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left({\sin }^{4}x\right).(1+log(\frac{2-sinx}{2+sinx}\left)\right).dx$

$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{4}x.(1-log(\frac{2+sinx}{2-sinx}\left)\right).dx$ ------$ii$

Adding equation i and ii we get,

$2I=2{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{4}x.dx$

$2I=4{\int }_0^{\frac{\pi }{2}}{\sin }^{4}x.dx$

$I=2{\int }_0^{\frac{\pi }{2}}{\sin }^{4}x.dx$

Use the reduction formula

${\int }^{}si{n}^{m}x.dx=\frac{-cosx.{\sin }^{m-1}\left(x\right)}{m}+\frac{m-1}{m}{\int }^{}{sin}^{-2+m}x.dx$

Here,$2{\int }_0^{\frac{\pi }{2}}si{n}^{4}x.dx=\frac{-2cosx.si{n}^{3}x}{4}{|}_0^{\frac{\pi }{2}}+\frac{3.2}{4}{\int }_0^{\frac{\pi }{2}}si{n}^{2}x.dx$ (As $\frac{-cosx.si{n}^{3}x}{}{|}_0^{\frac{\pi }{2}}=0$)

$\Rightarrow \frac{6}{4}{\int }_0^{\frac{\pi }{2}}\frac{1-cos2x}{2}.dx$

$=\frac{6}{8}{\int }_0^{\frac{\pi }{2}}1.dx-\frac{6}{8}{\int }_0^{\frac{\pi }{2}}cos2x.dx$

After putting the limits we get,

$=\frac{3\pi }{8}$ As$({\int }_0^{\frac{\pi }{2}}cos2x.dx)=0$

$\Rightarrow \frac{3\pi }{8}$