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Q.
If $\int \frac{ e ^{ x }(1+\sin x ) dx }{1+\cos x }= e ^{ x } f ( x )+ C$, then $f ( x )$ is equal to
Integrals
Solution:
$\int e ^{ x } \frac{(1+\sin x )}{(1+\cos x )} dx$
$=\int e ^{ x }\left[\frac{1}{2} \sec ^{2} \frac{ x }{2}+\tan \frac{ x }{2}\right] d x$
$=\frac{1}{2} \int e ^{ x } \sec ^{2} \frac{ x }{2} d x +\int e ^{ x } \tan \frac{ x }{2} dx$
$= e ^{ x } \tan \frac{ x }{2}+ C$
But $I = e ^{ x } f ( x )+ C \quad$ (given)
$\therefore f ( x )= \tan \frac{ x }{2}$