Question

Prove that $\int \frac{\tan x}{a^2+b^2{\tan }^{2}x}dx=\frac{\log ((a^2-b^2)\cos 2x+a^2+b^2)}{(a^2-b^2)}$.

Answer 1

$I=\int \frac{\tan x}{a^2+b^2{\tan }^{2}x}dx$
Put $u=\tan x\Rightarrow du={\sec }^{2}xdx$
$\therefore I=\int \frac{u}{(u^2+1)(a^2+b^2{u}^2)}du$
Put $s=u^2\Rightarrow ds=2udu.$
$\therefore I=\frac{1}{2}\int \frac{1}{(s+1)(a^2+b^{2}s)}ds.$
$=\frac{1}{2}\int (\frac{1}{s(a-b)(a+b)+(a-b)(a+b)}-\frac{b^2}{a^2(a-b)(a+b)+b^{2}s(a-b)(a+b)})ds$
$=\frac{\log \left((u^2+1)(a-b)(a+b)\right)}{2(a-b)(a+b)}-\frac{\log \left((a+b)(a-b)(a^2+b^2{u}^2)\right)}{2(a-b)(a+b)}$
$=\frac{\log ((a^2-b^2)\cos 2x+a^2+b^2)}{2(a^2-b^2)}$