$\int_{0}^{\frac{\pi}{2}}\left(e^{\sin x}-e^{\cos x}\right) d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} 	ext {Let } I &=\int_{0}^{\frac{\pi}{2}}\left(e^{\sin x}-e^{\cos x}ight) d x ....(1)\ &=\int_{0}^{\pi / 2} e^{\sin \left(\frac{\pi}{2}-xight)}-e^{\cos \left(\frac{\pi}{2}-xight)} \mathrm{dx} \ &=\int_{0}^{\pi / 2} e^{\cos x}-e^{\sin x} d x....(2) \end{aligned}$ Adding equation (1) \& (2) we get $\begin{array}{l} 2 I=\int_{0}^{\pi / 2}\left(e^{\sin x}-e^{\cos x}-e^{\cos x}-e^{\sin x}ight) d x \ 2 I=0 \Rightarrow I=0 \end{array}$

$\int_{0}^{\frac{\pi}{2}}\left(e^{\sin x}-e^{\cos x}\right) d x=$

Solution