# Problem with Indefinite Integral $\int\frac {\cos^4x}{\sin^3x} dx$

I'm stuck with this integral

$\int\frac {\cos^4x}{\sin^3x} dx$

which I rewrote as

$\int \csc^3x \cos^4xdx$

then after using the half angle formula twice for $\cos^4x$ I got this

$\frac 14\int \csc^3x (1+\cos(2x))(1+\cos(2x))dx$

then after solving those products I got these integrals

$\frac 14 \{\int \csc^3xdx+2\int \csc^3x \cos(2x)dx + \int \csc^3x \cos^2(2x)dx\}$

I do know how to solve the $\int \csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!

Let $t=\cos x$, then $$\int \frac{\cos^4 x}{\sin^3 x}dx=-\int \frac{t^4}{(1-t^2)^2}dt.$$ Can you proceed?
• Yeah I can proceed from there but I don't really understand what's going on with that substitution sorry if it is a silly question but what's going on behind it? I'm guessing you're using $sen^2 x = 1-cos^2 x$ on the denominator but i don't get how it got powered by 2, again sorry if it is a silly question or something! – CryoCodex Feb 8 '16 at 14:47
• Your guessing is right. Since $dt=-\sin x dx$, denominator except minus sign will be $\sin^4 x$. Then apply $\sin^2 x=1- \cos^2 x$. – choco_addicted Feb 8 '16 at 14:58
• To see how you might come to this substitution, note that the integrand is a product of an even power of $\cos x$ and an odd power of $\sin x$. – Travis Willse Dec 29 '18 at 2:07
Write $$\cos^4x=(1-\sin^2x)(1-\sin^2x)$$ and expand. Then split the fraction. You'll then need to integrate $$\csc^3x$$ , $$\csc x$$ and $$\sin x$$ all of which is standard