Question

Show that $\int \frac{\sin x+\cos x}{\sqrt{(1+\sin 2x)}}dx=x.$

Answer 1

Let $I=\int \frac{\sin x+\cos x}{\sqrt{\sin 2x+1}}dx=\int (\frac{\sin x}{\sqrt{\sin 2x+1}}+\frac{\cos x}{\sqrt{\sin 2x+1}})dx$
$I=I_1+I_2$
Where
$I_1=\int \frac{\sin x}{\sqrt{\sin 2x+1}}dx=\int \frac{\sin x\sqrt{2\sin x\cos x}}{{(\sin x+\cos x)}^2}dx$
$=\int \frac{\tan x{\sec }^{2}x}{{\tan }^{3}x+{\tan }^{2}x+\tan x+1}dx$
Put $t=\tan x\Rightarrow dt={\sec }^{2}xdx$
Therefore
$I_1=\int \frac{t}{t^3+t^2+t+1}dt=\int (\frac{t+1}{2(t^2+1)}-\frac{1}{2(t+1)})dt$
$=\frac{1}{2}\int \frac{t}{t^2+1}dt+\frac{1}{2}\int \frac{1}{t^2+1}dt-\frac{1}{2}\int \frac{1}{t+1}dt$
$=\frac{1}{2}{\tan }^{-1}t+\frac{1}{4}\log (t^2+1)-\frac{1}{2}\log (t+1)$
$=\frac{1}{2}x+\frac{1}{4}\log ({\tan x}^2+1)-\frac{1}{2}\log (\tan x+1)$
And
$I_2=\int \frac{\cos x}{\sqrt{\sin 2x+1}}dx=\int \frac{\cos x\sqrt{2\sin x\cos x+1}}{{\left(\sin x+\cos x\right)}^2}dx$
$=\int \frac{{\sec }^{2}x}{{\tan }^{3}x+{\tan }^{2}x+\tan x+1}dx$
Put $u=\tan x\Rightarrow du={\sec }^{2}xdx$
Therefore
$I_2=\int \frac{1}{u^3+u^2+u+1}du=\int (\frac{1-u}{2(u^2+1)}+\frac{1}{2(u+1)})du$
$=-\frac{1}{2}\int \frac{u}{u^2+1}du+\frac{1}{2}\int \frac{1}{u^2+1}du+\frac{1}{2}\int \frac{1}{u+1}du$
$=-\frac{1}{4}\log (u^2+1)+\frac{1}{2}{\tan }^{-1}u+\frac{1}{2}\log (t+1)$
$=-\frac{1}{4}\log ({\tan x}^2+1)+\frac{1}{2}x+\frac{1}{2}\log (\tan x+1)$
Hence
$I=\frac{1}{2}x+\frac{1}{2}x=x$