Question:
Evaluate the following integrals:
$\int \frac{\sin 2 x}{(a+b \cos 2 x)^{2}} d x$
Solution:
Assume $a+b \cos 2 x=t$
$d(a+b \cos 2 x)=d t$
$-2 b \sin 2 x d x=d t$
$\sin 2 x d x=\frac{-d t}{2 b}$
$\Rightarrow \frac{-1}{2 b} \int \frac{d t}{t^{2}}$
$\Rightarrow \frac{-1}{2 b} \int \frac{1}{t^{2}} d t$
$\Rightarrow \frac{-1}{2 b} \int t^{-2} \cdot d t$
$\Rightarrow \frac{t^{-1}}{2 b}+c$
But $t=a+b \cos 2 x$
$\Rightarrow \frac{1}{2 b(a+b \cos 2 x)}+C$
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