Evaluate the following integrals:

Solution:

$\operatorname{Let} I=\int \frac{\sin x}{\cos 2 x} d x$

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,

$I=\frac{1}{\sqrt{2}} \sin ^{-1}(\sqrt{2} \cos x)+C$

$\therefore I=\int \frac{\sin x}{\cos 2 x} d x=\frac{1}{\sqrt{2}} \sin ^{-1}(\sqrt{2} \cos x)+C$