Question

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

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

Answer 1

The correct option is C $\pi$

Given, $I={\int }_{-\pi }^{\pi }\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}dx$....(i)
$\Rightarrow I={\int }_{-\pi }^{\pi }\frac{e^{\sin (2\pi -x)}}{e^{\sin (2\pi -x)}+e^{-\sin (2\pi -x)}}dx$
$\Rightarrow I={\int }_{-\pi }^{\pi }\frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}}dx$....(ii)
On adding Eqs. (i) and (ii), we get
$2I={\int }_{-\pi }^{\pi }\frac{e^{\sin x}+e^{-\sin x}}{e^{\sin x}+e^{-\sin x}}dx$
$\Rightarrow 2I={\int }_{-\pi }^{\pi }1dx$
$\Rightarrow 2I={\left[x\right]}_{-}^{\pi }=\pi +\pi$
$\Rightarrow I=\pi$