Integrate $\displaystyle\int \dfrac{e^x}{1+\cos x}dx$
My Effort; I couldn't nothing.
Write $\cos(x)=\dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ and then it's all easy.
$$\cos2A=2\cos^2A-1\iff1+\cos2A=?$$
Now $\dfrac{d\tan(ax)}{dx}=a\sec^2(ax)$
$$\frac1{1+\cos x}=\frac{1-\cos x}{\sin^2x}$$
and now split the integral in two almost immediate ones.
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Write $\cos(x)=\dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ and then it's all easy.