Question

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

• A. $x\cot x.$
• B. $x\tan x.$
• C. $x^2\tan x.$
• D. $x\sec x.$

Answer 1

Answer: (B)

$x\tan x.$

Let $I=\int \frac{2x+\sin 2x}{\cos 2x+1}dx=\int (\frac{2x}{\cos 2x+1}+\frac{\sin 2x}{\cos 2x+1})dx$
$\Rightarrow I=I_1+I_2$
Where
$I_1=\int \frac{\sin 2x}{\cos 2x+1}dx$
Put $t=2x\Rightarrow dt=2dx$
$I_1=\frac{1}{2}\int \frac{\sin t}{\cos t+1}dt$
Now put $u=\cos t+1\Rightarrow du=-\sin tdt$
$I_1=-\frac{1}{2}\int \frac{du}{u}=-\frac{\log u}{2}$
$=-\frac{\log (\cos t+1)}{2}-\frac{\log (\cos 2x+1)}{2}$
And
$I_2=\int \frac{2x}{\cos 2x+1}dx=\int x{\sec }^{2}xdx$
$=x\tan x-\int \tan xdx=x\tan x+\log \left(\cos x\right)$
Hence
$I=-\frac{\log (\cos 2x+1)}{2}+x\tan x+\log \left(\cos x\right)$
$=x\tan x-\frac{1}{2}\log \left(2{\cos }^{2}x\right)+\log \left(\cos x\right)$
$=x\tan x+c$
Hence, option 'B' is correct.