Question:
Evaluate the following integrals:
$\int \frac{\sin x}{(1+\cos x)^{2}} d x$
Solution:
Assume $1+\cos x=t$
$\Rightarrow d(1+\cos x)=d t$
$\Rightarrow-\sin x \cdot d x=d t$
Substituting the values oft and dt we get
$\Rightarrow-\int \frac{\mathrm{dt}}{\mathrm{t}^{2}}$
$\Rightarrow-\int \frac{1}{\mathrm{t}^{2}} \mathrm{dt}$
$\Rightarrow-\int \mathrm{t}^{-2} \cdot \mathrm{dt}$
$\Rightarrow \frac{t^{-1}}{1}+c$
But $t=1+\cos x$
$\Rightarrow \frac{+1}{1+\cos x}+C$
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