Question

$\int {\sec }^{2}x\log (1+{\sin }^{2}x)dx=\tan x\log (1+{\sin }^{2}x)-2x+\sqrt{k}{\tan }^{-1}\sqrt{k}\tan x$. Find the value of $k$.

Answer 1

Integrating by parts, we get
$I=\tan x\log (1+{\sin }^{2}x)-\int \frac{\tan x}{1+{\sin }^{2}x}.2\sin x\cos xdx$ or $I=\tan x\log (1+{\sin }^{2}x)-2I_1$ ...(A)
where $I_1=\int \frac{{\sin }^{2}x}{1+{\sin }^{2}x}dx=\int (1-\frac{1}{1+{\sin }^{2}x})dx$
$=x-\int \frac{{\sec }^{2}xdx}{({\tan }^{2}x+1)+{\tan }^{2}x}$
$=x-\int \frac{dt}{(2t^2+1)}=x-\frac{1}{\sqrt{2}}{\tan }^{-1}t\sqrt{2}$
Putting the value of $I_1$ in (A), we get
$I=\tan x\log (1+{\sin }^{2}x)-2x+\sqrt{2}{\tan }^{-1}\sqrt{2}\tan x$