Question

Solve:

$\int \frac{{\cos }^{4}x-{\sin }^{4}x}{\sqrt{1+\sin 2x}}dx$

Answer 1

We have,

$I=\int \frac{{\cos }^{4}x-{\sin }^{4}x}{\sqrt{1+\sin 2x}}dx$

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

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

$I=\int \frac{\cos 2x}{\sqrt{1+\sin 2x}}dx$ $\because {\cos }^{2}x-{\sin }^{2}x=\cos 2x$

Let

$t=1+\sin 2x$

$\frac{dt}{dx}=0+2\cos 2x$

$\frac{dt}{2}=\cos 2xdx$

Therefore,

$I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$I=\frac{1}{2}\times 2\sqrt{t}+C$

$I=\sqrt{t}+C$

On putting the value of $t$, we get

$I=\sqrt{1+\sin 2x}+C$

Hence, this is the answer.