# Integral Substitution $\text{csc }\theta \cot\theta$

I have come across a question when researching integral substitution and I have no idea how to do it. Any help would be greatly appreciated as I don't understand what to do.

$$\int \text{csc}^2 2\theta \cot2\theta \,d\theta$$

I need to do the integral in two ways by using two different substitutions: $u = \cot 2\theta$ and $u = \text{csc } 2\theta$

• Why not write it out in terms of sine and cosine? – Mark Bennet Mar 29 '17 at 9:20
• @MarkBennet I don't know how to.. – Jessie Mar 29 '17 at 9:24
• @ShaunaGoodmanFitzpatrick: come on, don't you know what $\text{cosec}$ and $\cot$ are ??? – Yves Daoust Mar 29 '17 at 9:49

Solution 1: Let $u=\cot 2\theta$. Then $du=-2\csc^2 2\theta d\theta$. Thus,\begin{align} \int \text{csc}^2 2\theta \cot2\theta \,d\theta&=-\frac{1}{2}\int udu\\ &=-\frac{1}{2}\frac{u^2}{2}+k\\ &=-\frac{1}{2}\frac{\cot^2 2\theta}{2}+k \end{align}
Solution 2: Let $u=\csc 2\theta$. Then $du=-2\csc 2\theta \cot 2\theta d\theta$. Thus, \begin{align} \int \text{csc}^2 2\theta \cot2\theta \,d\theta&=\int \big[\csc 2\theta\cdot\csc 2\theta\cot 2\theta d\theta\big]\\ &=-\frac{1}{2}\int udu\\ &=-\frac{1}{2}\frac{u^2}{2}+C\\ &=-\frac{1}{2}\frac{\csc^22\theta}{2}+C\\ \end{align}