<s>[INST] 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?
 [/INST] HINT:
$$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx=\int \frac{(1-\cos^2(x))^2}{1+\cos^2(x)}\ dx$$
$$=\int \frac{(1+\cos^2(x))^2-4(1+\cos^2(x))+4}{1+\cos^2(x)}\ dx$$
$$=\int \left(1+\cos^2(x)-4+\frac{4}{1+\cos^2(x)}\right)\ dx$$
$$=\int \left(\frac{1+\cos(2x)}{2}-3+\frac{4\sec^2x}{\tan^2(x)+2}\right)\ dx$$
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