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Prove

Question:

$\frac{\sin x}{(1+\cos x)^{2}}$

Solution:

Let $1+\cos x=t$

$\therefore-\sin x d x=d t$

$\Rightarrow \int \frac{\sin x}{(1+\cos x)^{2}} d x=\int-\frac{d t}{t^{2}}$

$=-\int t^{-2} d t$

$=\frac{1}{t}+\mathrm{C}$

$=\frac{1}{1+\cos x}+\mathrm{C}$

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