Question
Integrate the function ${e^x}\left( {\frac{{1 + \sin x}}{{1 + \cos x}}} \right)$

Answer

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

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

$ = \int {{e^x}\left[ {\frac{1}{{2{{\cos }^2}\frac{x}{2}}} + \frac{{2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}} \right]dx} $

$= \int {{e^x}\left( {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right)dx}$

$= \int {{e^x}\left( {\tan \frac{x}{2} + \frac{1}{2}{{\sec }^2}\frac{x}{2}} \right)} dx$

$\left[ {\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx} \right]$

It is in the form of $\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}} dx$ since here $f\left( x \right) = \tan \frac{x}{2}$ and $f'\left( x \right) = \frac{1}{2}{\sec ^2}\frac{x}{2}$ 

$ = {e^x}\tan \frac{x}{2} + c$

$\left[ {\because \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx = {e^x}f\left( x \right) + c} } \right]$

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