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

Question

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!

Answer 1

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

Answer 2

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?

Answer 3

Integrate by parts instead

\begin{align} \int\frac {\cos^4x}{\sin^3x} dx =&-\frac12\int \cos^3x\ d\bigg(\frac1{\sin^2x}\bigg)\\ \overset{ibp}= &-\frac{\cos^3x}{2\sin^2x}+\frac32\int \left(\sin x-\frac{1}{\sin x}\right)dx\\ = &-\frac{\cos^3x}{2\sin^2x}-\frac32\cos x+\frac32\tanh^{-1}(\cos x) \end{align}