Let $I=\int \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x$
Dividing numerator and denominator by $\cos ^{4} x$, we get
$I=\int \frac{\tan x \sec ^{2} x}{\tan ^{4} x+1} d x$
Put $\tan ^{2} x=t \Rightarrow 2 \tan x \sec ^{2} x d x=d t$
$I=\frac{1}{2} \int \frac{d t}{1+t^{2}}=\frac{1}{2} \tan ^{-1} t=\frac{1}{2} \tan ^{-1}\left(\tan ^{2} x\right)+c$