Question

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

• A. $\frac{\sec A}{\sqrt{2}}\sqrt{\tan x\cos A-\sin A}+C$
• B. $\sqrt{2}\sec A\sqrt{\tan x\cos A-\sin A}+C$
• C. $\sqrt{2}\sec A\sqrt{\tan x\cos A+\sin A}+C$
• D. none of these

Answer 1

The correct option is C $\sqrt{2}\sec A\sqrt{\tan x\cos A+\sin A}+C$

$I=\int \frac{\sec xdx}{\sqrt{2\sin (x+A)\cos x}}$
$=\int \frac{{\sec }^{2}xdx}{\sqrt{\frac{2\sin (x+A)}{\cos x}}}$
$=\frac{1}{\sqrt{2}}\int \frac{{\sec }^{2}xdx}{\sqrt{\tan x\cos A+\sin A}}$
Put $\tan x\cos A+\sin A=t^2$
$\Rightarrow \cos A{\sec }^{2}xdx=2tdt$
$=\frac{\sec A}{\sqrt{2}}\int \frac{2tdt}{t}$
$=\sqrt{2}\sec A\int dt$
$I=\sqrt{2}\sec A\sqrt{\tan x\cos A+\sin A}+C$