Question

Evaluate $\int \frac{\tan x{\sec }^{2}x}{a+b{\tan }^{2}x}dx=$

Answer 1

$\int \frac{\tan x{\sec }^{2}x}{a+b{\tan }^{2}x}$

Let $\tan x=t$

On differentiating w.r.t $t,$ we have

$\Rightarrow {\sec }^{2}x.dx=dt$

$\int \frac{t}{a+b{t}^2}dt$

Let $a+b{t}^2=M$

Again differentiating w.r.t $t,$ we have

$\Rightarrow 2btdt=dM$

$\Rightarrow tdt=\frac{dM}{2b}$

$\Rightarrow \int \frac{dM}{2bCM}=\frac{1}{2b}=\frac{1}{2b}\log M+C=\frac{1}{2b}\log M+C=\frac{1}{2b}\log (a+b{t}^2)+C$

$=\frac{1}{2b}\log (a+b{\tan }^x)+C.$

Hence, solve.