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$\int_{0}^{\frac{\pi}{2}}\left(e^{\sin x}-e^{\cos x}\right) d x=$
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Verified Answer
The correct answer is:
$0$
$$
\text { Let } \begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}}\left(e^{\sin x}-e^{\cos x}\right) d x ....(1)\\
&=\int_{0}^{\pi / 2} e^{\sin \left(\frac{\pi}{2}-x\right)}-e^{\cos \left(\frac{\pi}{2}-x\right)} \mathrm{dx} \\
&=\int_{0}^{\pi / 2} e^{\cos x}-e^{\sin x} d x....(2)
\end{aligned}
$$
Adding equation (1) \& (2) we get
$$
\begin{array}{l}
2 I=\int_{0}^{\pi / 2}\left(e^{\sin x}-e^{\cos x}-e^{\cos x}-e^{\sin x}\right) d x \\
2 I=0 \Rightarrow I=0
\end{array}
$$
\text { Let } \begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}}\left(e^{\sin x}-e^{\cos x}\right) d x ....(1)\\
&=\int_{0}^{\pi / 2} e^{\sin \left(\frac{\pi}{2}-x\right)}-e^{\cos \left(\frac{\pi}{2}-x\right)} \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}\right) d x \\
2 I=0 \Rightarrow I=0
\end{array}
$$
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