Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easily solved using the substitution $y=\cos(x)$.
What if the power of sine is even?

Hint:
Bioche's rules suggest to use the substitution
$$t=\tan x,\quad \mathrm d x=\frac{\mathrm dt}{1+t^2}.$$
Indeed, as $\cos^2x=\frac 1{1+t^2}$, $\:\sin ^2x=\frac{t^2}{1+t^2}$, one obtains
$$\int\frac{\sin^4(x)}{1+\cos^2(x)}\,\mathrm dx=\int\frac{\frac{t^4}{(1+t^2)^2}}{1+\frac1{1+t^2}}\,\frac{\mathrm dt}{1+t^2}= \int\frac{t^4\,\mathrm dt}{(1+t^2)^2(2+t^2)}$$
which is comptuted with a decomposition into  partial fractions:
$$\frac{t^4}{(1+t^2)^2(2+t^2)}=\frac{At+B}{1+t^2}+\frac{Ct+D}{(1+t^2)^2}+\frac{Et+F}{2+t^2}.$$
