Question

The value of ${\int }_{-\pi /2}^{\pi /2}\frac{dx}{e^{\sin x}+1}$ is $k.$, then $2sin\left(k\right)=?$

Answer 1

Let $I={\int }_{-\pi /2}^{\pi /2}\frac{dx}{e^{\sin x}+1}={\int }_{-\pi /2}^0\frac{dx}{e^{\sin x}+1}+{\int }_0^{\pi /2}\frac{dx}{e^{\sin x}+1}$
$\Rightarrow I=I_1+I_2$
Put $x=-y$ in $I_1$
$I=-{\int }_{\pi /2}^0\frac{dy}{e^{-\sin y}+1}+{\int }_0^{\pi /2}\frac{dx}{1+e^{\sin x}}$
$\Rightarrow I={\int }_0^{\pi /2}\frac{e^{\sin x}dx}{e^{\sin x}+1}+{\int }_0^{\pi /2}\frac{dx}{e^{\sin x}+1}$
$={\int }_0^{\pi /2}1dx={\left[x\right]}_0^{\pi /2}=(\pi /2-0)=\pi /2$
$\therefore 2\sin \frac{\pi }{2}=2$