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?

Denote
$${I_n = \int\frac{\sin^{2n}(x)}{1 + \cos^2(x)}dx}$$
Then
$${I_{n}=\int\sin^2(x)\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx=\int(1-\cos^2(x))\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx}$$
If you expand this, you get
$${=I_{n-1} - \int \cos^2(x)\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx=I_{n-1}-\int \sin^{2n-2}(x) - \frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx}$$
Hence
$${I_{n}=2I_{n-1} - \int \sin^{2n-2}(x)dx}$$
Now define ${S_n = \int\sin^{2n}(x)dx}$. Then
$${S_{n}=\int \sin^2(x)\sin^{2n-2}(x)dx=S_{n-1}-\int \cos^2(x)\sin^{2n-2}(x)dx}$$
On the rightmost integral, using integration by parts yields
$${\int\cos^2(x) \sin^{2n-2}(x)dx=\frac{\cos(x)\sin^{2n-1}(x)}{2n-1}+\frac{1}{2n-1}\int \sin^{2n}(x)dx}$$
So overall
$${\Rightarrow S_n = S_{n-1}-\frac{\cos(x)\sin^{2n-1}(x)}{2n-1} - \frac{1}{2n-1}S_n}$$
And so
$${\left(\frac{2n}{2n-1}\right)S_n = S_{n-1} - \frac{\cos(x)\sin^{2n-1}(x)}{2n-1}}$$
$${\Rightarrow S_n = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$
Now you have two recursive relations that will help you compute the integral for higher even powers of ${\sin(x)}$:
$${I_n = 2I_{n-1} - S_{n-1}}$$
$${S_{n} = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$
