Question

${\int }_0^{\frac{\pi }{2}}\frac{\cos x}{1+e\sin x}dx$

Answer 1

We have,

$I={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{1+e\sin x}dx$

Let

$t=1+e\sin x$

$\frac{dt}{dx}=0+e\cos x$

$\frac{dt}{e}=\cos xdx$

Therefore,

$I=\frac{1}{e}{\int }_1^{1+e}\frac{1}{t}dt$

$I=\frac{1}{e}[\ln t{]}_1^{1+e}$

$I=\frac{1}{e}\left[\ln \right(1+e)-\ln 1]$

$I=\frac{\ln (1+e)}{e}$

Hence, this is the answer.