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?

$$\sin^4x=(\sin^2x+1)(\sin x+\sqrt{2})(\sin x-\sqrt{2})+(2+\sin^2x)$$
$$1+\cos^2x=(\sqrt{2}+\sin x)(\sqrt{2}-\sin x)$$
and so your integral becomes:
$$\int\frac{\sin^4x}{1+\cos^2x}dx=\int\frac{(\sin^2x+1)(\sin x+\sqrt{2})(\sin x-\sqrt{2})+(2+\sin^2x)}{(\sqrt{2}+\sin x)(\sqrt{2}-\sin x)}dx=-\int1+\sin^2xdx+\int\frac{2+\sin^2x}{2-\sin^2x}dx$$
