<s>[INST] Integrate $\int e^x\frac{1+\sin x}{1+\cos x}dx$ Integrate
$$\int e^x\cdot\frac{1+\sin x}{1+\cos x}\,dx$$
My try;
First step:
I let
$$\frac{1+\sin x}{1+\cos x} = u$$
$$e^x = v$$
and then I applied integration by parts:
$$\frac{1+\sin x}{1+\cos x}=u \implies du=\frac{\sin x+\cos x+1}{(1+\cos x)^2}dx$$
$$v = e^x \implies dv = e^x$$
and
$$\int e^x\cdot\frac{1+\sin x}{1+\cos x}\,dx=e^x\cdot\frac{1+\sin x}{1+\cos x}-\int e^x\cdot\frac{\sin x+\cos x+1}{(1+\cos x)^2}dx$$
 [/INST] Use $\sin x=2\sin\frac{x}{2}\cos\frac{x}{2},1+\cos x=2\cos^2\frac{x}{2}$. The integrand becomes $e^x\tan\frac{x}{2}+\frac{1}{2}e^x\sec^2\frac{x}{2}$.
Since the derivative of $\tan\frac{x}{2}$ is $\frac{1}{2}\sec^2\frac{x}{2}$ we can integrate immediately to get $e^x\tan\frac{x}{2}+C$.
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