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?

  • $\begingroup$ 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! $\endgroup$ – CryoCodex Feb 8 '16 at 14:47
  • $\begingroup$ Your guessing is right. Since $dt=-\sin x dx $, denominator except minus sign will be $\sin^4 x $. Then apply $\sin^2 x=1- \cos^2 x$. $\endgroup$ – choco_addicted Feb 8 '16 at 14:58
  • $\begingroup$ Ohh i got it now! haha thanks man! $\endgroup$ – CryoCodex Feb 8 '16 at 15:03
  • $\begingroup$ 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 $\sin x$. $\endgroup$ – 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

  • $\begingroup$ I'll try doing it this way too thanks man! $\endgroup$ – CryoCodex Feb 8 '16 at 15:12

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