Evaluate the following integrals:

Given $I=\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x$

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

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

Putting $\tan ^{2} x=t$ so that $2 \tan x \sec ^{2} x d x=d t$

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

We know that $\int \frac{1}{1+x^{2}} d x=\tan ^{-1}(x)+c$

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\tan ^{-1}(\mathrm{t})+\mathrm{c}$

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

$\therefore \mathrm{I}=\int \frac{\sin 2 \mathrm{x}}{\sin ^{4} \mathrm{x}+\cos ^{4} \mathrm{x}} \mathrm{dx}=\tan ^{-1}\left(\tan ^{2} \mathrm{x}\right)+\mathrm{c}$