Question

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

• A. $\frac{3\pi }{2}$
• B. $\frac{\pi }{2}$
• C. $\frac{5\pi }{4}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is D $\frac{3\pi }{4}$

$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{1+{\sin }^{2}x}{1+{\pi }^{\sin x}}\right)dx\cdots \left(i\right)$
Applying property of definite integrals
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{1+{\sin }^{2}x}{1+{\pi }^{-\sin x}}\right)dx$
$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{{\pi }^{\sin x}(1+{\sin }^{2}x)}{(1+{\pi }^{\sin x})}dx\cdots \left(ii\right)$
Adding equation $\left(i\right)$ and $\left(ii\right)$
$2I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{(1+{\sin }^{2}x)(1+{\pi }^{\sin x})}{(1+{\pi }^{\sin x})}dx$
$2I=\underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}(1+{\sin }^{2}x)dx$
$2I=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}(1+{\sin }^{2}x)dx$
$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}1+\frac{1-\cos 2x}{2}dx$
$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}(\frac{3}{2}-\frac{1}{2}\cos 2x)dx={[\frac{3}{2}x-\frac{1}{4}\sin 2x]}_0^{\frac{\pi }{2}}$
$I=\frac{3}{2}\left(\frac{\pi }{2}\right)-\frac{1}{4}\left(\sin \pi \right)$
$\therefore I=\frac{3\pi }{4}$