Question

If $\int \frac{1}{1-{\sin }^{4}x}dx=\frac{1}{a\sqrt{b}}{\tan }^{-1}\left(\sqrt{2}\tan x\right)+\frac{1}{b}\tan c+C$ then find $\frac{a}{b}$

Answer 1

$\int \frac{1}{\left(1-{\sin }^{4}x\right)}dx=\int \frac{{\sec }^{4}xdx}{\left({\sec }^{4}x-{\tan }^{4}x\right)}$

Put

$\tan x=t$

${\sec }^{2}xdx=dt$

Therefore,

$=\int \frac{\left(1+t^2\right)dt}{\left\{{\left(1+t^2\right)}^2-t^4\right\}}=\int \frac{\left(1+t^2\right)dt}{\left(t^4+2t^2+1-t^4\right)}=\int \frac{\left(1+t^2\right)dt}{\left(2t^2+1\right)}=\int \frac{dt}{2t^2+1}+\int \frac{t^2}{2t^2+1}dt=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}t\right)+\frac{1}{2}\int \frac{\left(2t^2+1-1\right)dt}{\left(2t^2+1\right)}=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\tan x\right)+\frac{1}{2}\tan x-\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\tan x\right)+C=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\tan x\right)+\frac{1}{2}\tan x+C\therefore a=2,b=2$

$So,\frac{a}{b}=1$

$Hence,solved.$