Question

$\int \frac{\sec x}{\sqrt{\sin (2x+\alpha )+\sin \alpha }}dx$ is equal to

• A. $\sqrt{2\sec \alpha (\tan x+\tan \alpha )}+c$
• B. $\sqrt{2\sec \alpha (\tan x-\tan \alpha )}+c$
• C. $\sqrt{2\sec \alpha (\tan \alpha -\tan x)}+c$
• D. $\sqrt{2\tan \alpha (\sec x-\sec \alpha )}+c$

Answer 1

The correct option is A $\sqrt{2\sec \alpha (\tan x+\tan \alpha )}+c$

$\int \frac{\sec x}{\sqrt{\sin (2x+\alpha )+\sin \alpha }}dx$
$=\int \frac{\sec x}{\sqrt{\sin 2x\cos \alpha +\cos 2x\sin \alpha +\sin \alpha }}dx$
$=\int \frac{\sec x}{\sqrt{\sin 2x\cos \alpha +(\cos 2x+1)\sin \alpha }}dx$
$=\int \frac{\sec x}{\sqrt{2\cos x(\sin x\cos \alpha +\cos x\sin \alpha )}}dx$
$=\int \frac{\sec x}{\sqrt{2{\cos }^{2}x\cos \alpha (\tan x+\tan \alpha )}}dx$
$=\frac{1}{\sqrt{2\cos \alpha }}\int \frac{{\sec }^{2}x}{\sqrt{(\tan x+\tan \alpha )}}dx$
Put, $\tan x+\tan \alpha =t$

Differentiating above equation w.r.t. x,we get
$\Rightarrow {\sec }^{2}xdx=dt$
$=\frac{1}{\sqrt{2\cos \alpha }}\int \frac{dt}{\sqrt{t}}$
$=\frac{1}{\sqrt{2\cos \alpha }}\frac{t^{1/2}}{\frac{1}{2}}+C$
$=\sqrt{\frac{2}{\cos \alpha }}\sqrt{\tan x+\tan \alpha }+C$

$\sqrt{2\sec \alpha (\tan x+\tan \alpha )}+c$