Question

Evaluate $\int \frac{1}{si{n}^{4}x+co{s}^{4}x}dx$

Answer 1

$\int \frac{1}{{\sin }^{4}x+{\cos }^{4}x}dx$

Dividing numerator and denominator by ${\cos }^{4}x$, we have

$=\int \frac{\left(\frac{1}{{\cos }^{4}x}\right)}{\left(\frac{{\sin }^{4}x}{{\cos }^{4}x}\right)+\left(\frac{{\cos }^{4}x}{{\cos }^{4}x}\right)}dx$

$=\int \frac{{\sec }^{4}x}{1+{\tan }^{4}x}dx$

$=\int \frac{{\sec }^{2}x(1+{\tan }^{2}x)}{1+{\tan }^{4}x}dx$

Let

$\tan x=u$

$\Rightarrow {\sec }^{2}xdx=du$

$\therefore =\int \frac{1+u^2}{1+u^4}du$

$\Rightarrow =\int \frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}}du$

$\Rightarrow =\int \frac{1+\frac{1}{u^2}}{{(u-\frac{1}{u})}^2+2}du$

Now again,

Let

$u-\frac{1}{u}=t$

$\Rightarrow (1+\frac{1}{u^2})du=dt$

$\therefore =\int \frac{dt}{t^2+{\left(\sqrt{2}\right)}^2}$

$\Rightarrow =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)+C$

$\Rightarrow =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{(u-\frac{1}{u})}{\sqrt{2}}\right)+C[\because t=u-\frac{1}{u}]$

$\Rightarrow =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x-\frac{1}{\tan x}}{\sqrt{2}}\right)+C[\because u=\tan x]$

$\Rightarrow =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{{\tan }^{2}x-1}{\sqrt{2}\tan x}\right)+C$

Hence $\int \frac{1}{{\sin }^{4}x+{\cos }^{4}x}dx=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{{\tan }^{2}x-1}{\sqrt{2}\tan x}\right)+C$