Integrating $ 3 \int \frac{\cos^5 x}{\sin x} dx$?

Question

I need to solve the integral $$3 \int \frac{\cos^5 x}{\sin x} dx$$

I'm not sure how to proceed from here. I think Integration by Parts may be useful, but I'm not entirely sure what to make $u$ and $dv$.

Answer 1

A start: Note that $\cos^4x=(1-\sin^2 x)^2$. Let $u=\sin x$.

Answer 2

with $t=\cos(x)$ we get $dt=-\sin(x)dx$ and we obtain $-3 \int \frac{t^5}{\sin(x)^2}dt=-3\int \frac{t^5}{1-t^2}dt$
Sonnhard.

Answer 3

$$ \int \frac{\cos^5x}{\sin x}\,dx = \int\frac{\cos^4x \cdot \cos x}{\sin x}\,dx $$ $$ = \int \frac{(1-\sin^2 x)^2 \cos x}{\sin x}\, dx $$ Then put $t=\sin x$ then $dt=\cos x\, dx$

Then the integral is equal to $$\int\frac{(1-t^2)^2}{t}\, dt$$

Answer 4

Hint:

To integrate such trigonometric fractions , you have to look at the degrees. If the integrand is of the form $\sin x\,F(\cos x)/G(\cos x)$ or $\cos x\,F(\sin x)/G(\sin x)$, the change of variable to an ordinary rational fraction is easy.

The cases of $\sin x\,F(\cos x,\sin^2x)/G(\cos x,\sin^2x)$ and $\cos x\,F(\sin x,\cos^2x)/G(\sin x,\cos^2x)$ are handled similary by using the identity $\sin^2x+\cos^2x=1$.