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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 $u = \text{csc } 2\theta$

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

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, $$\begin{align} \int \text{csc}^2 2\theta \cot2\theta \,d\theta&=\int \big[\csc 2\theta\cdot\csc 2\theta\cot 2\theta d\theta\big]\\ &=-\frac{1}{2}\int udu\\ &=-\frac{1}{2}\frac{u^2}{2}+C\\ &=-\frac{1}{2}\frac{\csc^22\theta}{2}+C\\ \end{align}$$

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  • $\begingroup$ I need to use the two substitutions given though $\endgroup$ – Jessie Mar 29 '17 at 9:25
  • $\begingroup$ Can you please explain where you got x = 2θ? $\endgroup$ – Jessie Mar 29 '17 at 9:38
  • $\begingroup$ On second thoughts I discarded my answer, as the question explicitly asks for these changes of variable. $\endgroup$ – Yves Daoust Mar 29 '17 at 9:58
  • $\begingroup$ Do you know of any good sources I could learn this about? I don't understand where most of these lines are coming from $\endgroup$ – Jessie Mar 29 '17 at 9:58
  • $\begingroup$ @Shauna Goodman Fitzpatrick Just goggle about trigonometric integrals and you will a lot of examples. $\endgroup$ – Juniven Mar 29 '17 at 10:01

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