# How do you evaluate int cot^5(2x) dx?

Apr 19, 2018

$I = \frac{1}{8} \left(4 \ln \left(\sin \left(2 x\right)\right) - {\csc}^{4} \left(2 x\right) + 4 {\csc}^{2} \left(2 x\right)\right) + C$

#### Explanation:

We want to integrate

$I = \int {\cot}^{5} \left(2 x\right) \mathrm{dx}$

Make a substitution color(red)(u=2x=>du=2dx

$I = \frac{1}{2} \int {\cot}^{5} \left(u\right) \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{2} \int \frac{{\left({\cos}^{2} \left(u\right)\right)}^{2} \cos \left(u\right)}{\sin} ^ 5 \left(u\right) \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{2} \int \frac{{\left(1 - {\sin}^{2} \left(u\right)\right)}^{2} \cos \left(u\right)}{\sin} ^ 5 \left(u\right) \mathrm{du}$

Make a substitution color(red)(s=sin(u)=>ds=cos(u)du

$I = \frac{1}{2} \int {\left(1 - {s}^{2}\right)}^{2} / {s}^{5} \mathrm{ds}$

$\textcolor{w h i t e}{I} = \frac{1}{2} \int \frac{1}{s} + \frac{1}{s} ^ 5 - 2 \frac{1}{s} ^ 3 \mathrm{ds}$

$\textcolor{w h i t e}{I} = \frac{1}{2} \left(\ln \left(s\right) - \frac{1}{4 {s}^{4}} + \frac{1}{s} ^ 2\right) + C$

$\textcolor{w h i t e}{I} = \frac{1}{8} \left(4 \ln \left(s\right) - \frac{1}{s} ^ 4 + \frac{4}{s} ^ 2\right) + C$

Substitute back color(red)(s=sin(u) and color(red)(u=2x

$I = \frac{1}{8} \left(4 \ln \left(\sin \left(2 x\right)\right) - {\csc}^{4} \left(2 x\right) + 4 {\csc}^{2} \left(2 x\right)\right) + C$