# Integrate csc^3 2x dx ?

Apr 4, 2018

$I = \frac{1}{16} \left({\tan}^{2} \left(x\right) + 4 \ln \left(\tan \left(x\right)\right) - \frac{1}{{\tan}^{2} \left(x\right)}\right) + C$

#### Explanation:

We want to solve

$I = \int {\csc}^{3} \left(2 x\right) \mathrm{dx}$

Make a substitution $u = 2 x \implies \mathrm{du} = 2 \mathrm{dx}$

$I = \frac{1}{2} \int {\csc}^{3} \left(u\right) \mathrm{du}$

Use tangent half-angle substitution $s = \tan \left(\frac{u}{2}\right)$,
then $\csc \left(u\right) = \frac{1 + {s}^{2}}{2 s}$ and $\mathrm{du} = \frac{2}{1 + {s}^{2}} \mathrm{ds}$

$I = \frac{1}{2} \int {\left(\frac{1 + {s}^{2}}{2 s}\right)}^{3} \frac{2}{1 + {s}^{2}} \mathrm{ds}$

$\textcolor{w h i t e}{I} = \int {\left(1 + {s}^{2}\right)}^{2} / \left(8 {s}^{3}\right) \mathrm{ds}$

$\textcolor{w h i t e}{I} = \frac{1}{8} \int \frac{{s}^{4} + 2 {s}^{2} + 1}{{s}^{3}} \mathrm{ds}$

$\textcolor{w h i t e}{I} = \frac{1}{8} \int s + 2 {s}^{-} 1 + {s}^{-} 3 \mathrm{ds}$

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

$\textcolor{w h i t e}{I} = \frac{1}{16} \left({s}^{2} + 4 \ln \left(s\right) - \frac{1}{{s}^{2}}\right) + C$

Substitute back $s = \tan \left(\frac{u}{2}\right)$ and $u = 2 x$

$I = \frac{1}{16} \left({\tan}^{2} \left(x\right) + 4 \ln \left(\tan \left(x\right)\right) - \frac{1}{{\tan}^{2} \left(x\right)}\right) + C$

Apr 4, 2018

$I = - \frac{1}{4} \left[\csc \left(2 x\right) \cot \left(2 x\right) + \ln | \csc \left(2 x\right) + \cot \left(2 x\right) |\right] + c$

#### Explanation:

Here,

$I = \int {\csc}^{3} 2 x \mathrm{dx}$

$= \int \left(\csc 2 x\right) \left({\csc}^{2} \left(2 x\right)\right) \mathrm{dx}$

$= \int \left(\sqrt{{\cot}^{2} \left(2 x\right) + 1}\right) {\csc}^{2} \left(2 x\right) \mathrm{dx}$

Let,

$\textcolor{b l u e}{\cot \left(2 x\right) = u} \implies - {\csc}^{2} \left(2 x\right) \cdot 2 \mathrm{dx} = \mathrm{du}$

$\implies {\csc}^{2} \left(2 x\right) \mathrm{dx} = - \frac{1}{2} \mathrm{du}$

$I = \int \sqrt{{u}^{2} + 1} \left(- \frac{1}{2}\right) \mathrm{du}$

We know that,

color(red)(intsqrt(x^2+a^2)dx=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+c

$I = - \frac{1}{2} \int \sqrt{{u}^{2} + {1}^{2}} \mathrm{du}$

$= - \frac{1}{2} \left[\frac{u}{2} \sqrt{{u}^{2} + 1} + \frac{1}{2} \ln | u + \sqrt{{u}^{2} + 1} |\right] + c$

substituting back color(blue)(u=cot2x

=-1/2[cot(2x)/2sqrt(cot^2(2x)+1)+1/2ln|cot(2x)+sqrt(cot^2(2x)+1 )|]+c

$= - \frac{1}{2} \left[\cot \frac{2 x}{2} \csc \left(2 x\right) + \frac{1}{2} \ln | \cot \left(2 x\right) + \csc \left(2 x\right) |\right] + c$

$= - \frac{1}{4} \left[\csc \left(2 x\right) \cot \left(2 x\right) + \ln | \csc \left(2 x\right) + \cot \left(2 x\right) |\right] + c$