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Integrate the function
$\frac{\sin x}{(1+\cos x)^2}$
$\frac{\sin x}{(1+\cos x)^2}$
Solution:
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Verified Answer
Put $1+\cos x=t$, so that $-\sin x d x=d t$
$\begin{aligned}
&\therefore \quad \int \frac{\sin x}{(1+\cos x)^2} d x=-\int \frac{d t}{t^2} \\
&=\frac{1}{t}+C=\frac{1}{1+\cos x}+C
\end{aligned}$
$\begin{aligned}
&\therefore \quad \int \frac{\sin x}{(1+\cos x)^2} d x=-\int \frac{d t}{t^2} \\
&=\frac{1}{t}+C=\frac{1}{1+\cos x}+C
\end{aligned}$
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