Evaluate the following integrals:


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


Consider $\int \frac{1}{\sin ^{4} x+\cos ^{4} x} d x$,

Divide num and denominator by $\cos ^{4} x$ to get,

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

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

$=\int \frac{\sec ^{2} x \cdot \sec ^{2} x}{\tan ^{4} x+1} d x$

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

Let $\tan x=t$

$\sec ^{2} x d x=d t$

$=\int \frac{\left(1+t^{2}\right)}{t^{4}+1} d t$

Now divide both numerator and denominator by $\frac{1}{t^{2}}$ to get,

$=\int \frac{\left(\frac{1}{t^{2}}+1\right)}{\left(t^{2}+\frac{1}{t^{2}}\right)+2-2} d t$

$=\int \frac{\left(\frac{1}{t^{2}}+1\right)}{\left(1-\frac{1}{t}\right)^{2}+2} d t$

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

$\left(1+\frac{1}{t^{2}}\right) d t=d u$

$=\int \frac{d u}{u^{2}+2}$

$=\int \frac{d u}{u^{2}+(\sqrt{2})^{2}}$

$=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+c$

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

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

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