# How do you integrate cscxcot^4xdx

May 4, 2015

$\int \csc \left(x\right) \cdot {\cot}^{4} \left(x\right) \mathrm{dx}$

Just trigonometric function : let's $t = \tan \left(\frac{1}{2} x\right)$

$\csc \left(x\right) = \frac{1 + {t}^{2}}{2 t}$

$\cot \left(x\right) = \frac{1 - {t}^{2}}{2 t}$

$\mathrm{dx} = \frac{2}{1 + {t}^{2}} \mathrm{dt}$

So now we have :

$\int \frac{1 + {t}^{2}}{2 t} \cdot {\left(\frac{1 - {t}^{2}}{2 t}\right)}^{4} \cdot \frac{2}{1 + {t}^{2}} \mathrm{dt}$

$\implies \int \frac{1}{t} \cdot {\left(\frac{1 - {t}^{2}}{2 t}\right)}^{4} \mathrm{dt}$

Use the Binomial theorem

$\implies \int \frac{1}{t} \cdot \frac{1 - 4 {t}^{2} + 6 {t}^{4} - 4 {t}^{6} + {t}^{8}}{16 {t}^{4}} \mathrm{dt}$

$\implies \frac{1}{16} \int \frac{1 - 4 {t}^{2} + 6 {t}^{4} - 4 {t}^{6} + {t}^{8}}{{t}^{5}} \mathrm{dt}$

$\implies \frac{1}{16} \int \frac{1}{t} ^ 5 - \frac{4}{t} ^ 3 + \frac{6}{t} - 4 t + {t}^{3} \mathrm{dt}$

$\implies \frac{1}{16} \left[- \frac{1}{4 {t}^{4}} + \frac{2}{t} ^ 2 + 6 \ln \left(| t |\right) - 2 {t}^{2} + \frac{1}{4} {t}^{4}\right] + C$

And then substitute back for $t = \tan \left(\frac{1}{2} x\right)$...