Question

If $I={\int }_{-\pi /2}^{\pi /2}\frac{dx}{e^{\sin x}+1}$ then $I$ equals

• A. $3\pi /2$
• B. $\pi$
• C. $\pi /2$
• D. $3\pi /4$

Answer 1

The correct option is C $\pi /2$

Let $I={\int }_{-\pi /2}^{\pi /2}\frac{dx}{e^{\sin x}+1}$ ...(1)
$={\int }_{-\pi /2}^{\pi /2}\frac{dx}{e^{\sin (\frac{\pi }{2}-\frac{\pi }{2}-x)}+1}$
$I={\int }_{-\pi /2}^{\pi /2}\frac{dx}{e^{-\sin x}+1}={\int }_{-\pi /2}^{\pi /2}\frac{e^{\sin x}dx}{e^{\sin x}+1}$ ...(2)
From (1) and (2), we get
$2I={\int }_{-\pi /2}^{\pi /2}\frac{e^{\sin x}+1}{e^{\sin x}+1}dx={\int }_{-\pi /2}^{\pi /2}dx=\pi$
$\Rightarrow I=\frac{\pi }{2}$