Question

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

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

Answer 1

The correct option is A $\frac{\pi }{2}$

$I=\underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{1+e^{\sin x}}dx\cdots \left(i\right)$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$I=\underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{1+e^{-\sin x}}dx$
$I=\underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{e^{\sin x}}{1+e^{\sin x}}dx\cdots \left(ii\right)$
Adding $\left(i\right),\left(ii\right)$
$\Rightarrow 2I=\underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{1+e^{\sin x}}{1+e^{\sin x}}dx$
$\Rightarrow 2I=[x{]}_{-\pi /2}^{\pi /2}$
$\Rightarrow 2I=\pi$
$\Rightarrow I=\frac{\pi }{2}$