How to find $\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\cot^5x\,\csc^3xdx$ 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 !
 A: 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 
$$\begin{align*}
\int(\csc^2 x-1)^2\csc^2 x(\csc x\cot x)dx&=-\int(u^2-1)^2u^2du\\
&=-\int(u^6-2u^4+u^2)du\\
&=-\frac{u^7}7+\frac{2u^5}5-\frac{u^3}3+C\\
&=-\frac17\csc^7 x+\frac25\csc^5 x-\frac13\csc^3 x+C\;.
\end{align*}$$
A: Here's one way to find the indefinite integral.
Write everything in terms of $\sin$ and $\cos$.  If one of those functions is raised to an odd positive power, say $\cos$, factor one term of $\cos$ out and use the Pythagorean identity to write the other factor in terms of $\sin$. Then use a $u$-substitution with $u=\sin x$:
$$\eqalign{
\int\cot^5 x\,\csc^3 x\,dx&= 
\int {\cos^5 x\over \sin^5 x}{1\over \sin^3 x}\,dx\cr
&=\int {\cos  x \cdot(\cos^2 x)^2\over \sin^8 x} \,dx\cr 
&=\int {\cos  x \cdot(1-\sin^2 x)^2\over \sin^8 x} \,dx\cr 
&\buildrel{u=\sin x}\over=\int {  (1-u^2)^2\over u^8  } \,du\cr 
&=\int {  1-2u^2+u^4 \over u^8  } \,du\cr 
&=\int {( u^{-8}-2u^{-6}+u^{-4}  )} \,du\cr 
&=  {{ -u^{-7}\over7}+2{u^{-5}\over 5}-{u^{-3}\over3}  } +C\cr 
&=  {{ -\sin^{-7} x\over7}+2{\sin^{-5}x\over 5}-{\sin^{-3}x\over3}  } +C.\cr 

}
$$
