I stack about following problem…
$\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\cot^5x\,\csc^3xdx$
I tried to change $\cot^5x=\frac{\cos^5x}{\sin^5x}$
I got
$$\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{\cos^5x}{\sin^5x}\csc^3xdx$$
but after that I couldn't get good idea to integral this function.
Anyone has any idea for the problem
Thank you !
Best Answer
Note that it’s not necessary to convert to sines and cosines:
$$\cot^5 x\,\csc^3 x=\cot^4x\,\csc^2 x(\cot x\csc x)= (\csc^2 x-1)^2\csc^2 x(\csc x\cot x)\;,$$
and $d(\csc x)=-\csc x\cot x dx$, so you can simply let $u=\csc x$. The indefinite integral then becomes
$$\begin{align*} \int(\csc^2 x-1)^2\csc^2 x(\csc x\cot x)dx&=-\int(u^2-1)^2u^2du\\ &=-\int(u^6-2u^4+u^2)du\\ &=-\frac{u^7}7+\frac{2u^5}5-\frac{u^3}3+C\\ &=-\frac17\csc^7 x+\frac25\csc^5 x-\frac13\csc^3 x+C\;. \end{align*}$$