Question

Evaluate $\int \frac{\cos 2x}{\sqrt{1+\sin 2x}}dx$.

Answer 1

We need to evaluate $\int \frac{cos2x}{\sqrt{1+sin2x}}dx$

Consider $\int \frac{cos2x}{\sqrt{1+sin2x}}dx$

Let $u=1+sin2x$

$\Rightarrow du=2cos2xdx$

$\Rightarrow cos2xdx=\frac{du}{2}$

$\therefore \int \frac{cos2x}{\sqrt{1+sin2x}}dx=\int \frac{du}{2\sqrt{u}}$

$=\frac{1}{2}\int \frac{du}{\sqrt{u}}$

$=\frac{1}{2}\left[\frac{u^{1/2}}{\left(1/2\right)}\right]$

$=u^{1/2}$

$=\sqrt{u}$

Substitute back $u=1+sin2x$ we get

$=\sqrt{1+sin2x}+c$ where $c$ is integration constant.

Thus $\int \frac{cos2x}{\sqrt{1+sin2x}}dx=\sqrt{1+sin2x}+c$