Evaluate $\int \frac{\sin x}{\cos 2 x} d x$
$\operatorname{Let} I=\int \frac{\sin x}{\cos 2 x} d x$
We know $\cos 2 x=2 \cos ^{2} x-1$
Putting values in I we get,
$I=\int \frac{\sin x}{\cos 2 x} d x=\int \frac{\sin x}{2 \cos ^{2} x-1} d x$
Put $\cos x=t$
Differentiating w.r.t to $x$ we get,
$\sin x d x=-d t$
Putting values in integral we get,
$I=-\int \frac{d t}{2 t^{2}-1}=-\int \frac{d t}{(\sqrt{2} t)^{2}-(1)^{2}}$
Again put $\sqrt{2} \times \mathrm{t}=\mathrm{u}$
Differentiating w.r.t to t we get,
$d t=\frac{d u}{\sqrt{2}}$
Putting values in integral we get,
$I=\frac{1}{\sqrt{2}} \int \frac{d u}{(1)^{2}-(u)^{2}}$
We know $\int \frac{d x}{(1)^{2}-(x)^{2}}=\sin ^{-1} x+C$
$I=\frac{1}{\sqrt{2}} \sin ^{-1} u+C$Substituting value of $u$ we get,
$I=\frac{1}{\sqrt{2}} \sin ^{-1} \sqrt{2} t+C$
Substituting value of $t$ we get,
$I=\frac{1}{\sqrt{2}} \sin ^{-1}(\sqrt{2} \cos x)+C$
$\therefore I=\int \frac{\sin x}{\cos 2 x} d x=\frac{1}{\sqrt{2}} \sin ^{-1}(\sqrt{2} \cos x)+C$