Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?

Question

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?

Answer 1

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: $$\frac{t^4}{(1+t^2)^2(2+t^2)}=\frac{At+B}{1+t^2}+\frac{Ct+D}{(1+t^2)^2}+\frac{Et+F}{2+t^2}.$$

Answer 2

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 ${\sin(x)}$:

$${I_n = 2I_{n-1} - S_{n-1}}$$

$${S_{n} = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$

Answer 3

HINT: $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx=\int \frac{(1-\cos^2(x))^2}{1+\cos^2(x)}\ dx$$ $$=\int \frac{(1+\cos^2(x))^2-4(1+\cos^2(x))+4}{1+\cos^2(x)}\ dx$$ $$=\int \left(1+\cos^2(x)-4+\frac{4}{1+\cos^2(x)}\right)\ dx$$ $$=\int \left(\frac{1+\cos(2x)}{2}-3+\frac{4\sec^2x}{\tan^2(x)+2}\right)\ dx$$

Answer 4

HINT

You can use $\sin(x) = \tan(x) / \sec(x)$ and $\cos(x) = 1/\sec(x)$.

and then it will bring you to Partial Fractions Decomposition.

Answer 5

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

Answer 6

If you enjoy special functions, using $t=\tan(x)$ $$I_n=\int \frac{\sin^n(x)}{1+\cos^2(x)}\ dx=\int \left(\frac{t}{\sqrt{t^2+1}}\right)^n\frac{dt}{t^2+2}$$

$$I_n=\frac {t^{n+1}}{2(n+1)}\,F_1\left(\frac{n+1}{2};\frac{n}{2},1;\frac{n+3}{2};-t^2,-\frac{t^2}{2}\right)$$ where appears the Appell hypergeometric function of two variables.