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How do you integrate $\csc^4 x/\cot^2 x$? I know that this is the same as $\csc^4 x \cot^{-2} x$ and when you use techniques in trig integrals you end up with $$\int \csc^2 x \csc^2 x \cot^{-2} x \,dx = \int \left(1 + \cot^2 x\right)\left(\cot^{-2} x\right)\csc^2 x \,dx.$$ Making the substitution $u = \cot x \Rightarrow du = -\csc^2 x \,dx$, $$\begin{align} \ldots{} &= -\int \left(1 + u^2 \right) u^{-2} \,du \\ &= -\int \left( u^{-2} + u^0 \right) \,du \\ &= -\int \left( u^{-2} + 1 \right) \,du \\ &= -\left[ -u^{-1} + u \right] \\ &= (1/\cot x) - \cot x \,. \end{align}$$ Is this the final answer or did I just mess up?

nivk
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sheila
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    Consider writing the question using mathjax. See here. – Bman72 Jan 07 '15 at 20:08
  • i'm sorry i'm really bad at codes – sheila Jan 07 '15 at 20:17
  • You should try much harder then, @sheila, otherwise chances are many people won't even take the time to try to read what you wrote. Try the following: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-qu%E2%80%8C%E2%80%8Bick-reference – Timbuc Jan 07 '15 at 20:20
  • ok i'll try i'm really sorry – sheila Jan 07 '15 at 20:22
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    A bit hard to read, but the procedure, work look right. It looks as if you forgot the $+{}C$. You could check correctness by differentiating, and playing a bit with trig identities that you have already used. – André Nicolas Jan 07 '15 at 20:29

2 Answers2

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Another approach is to use

$\displaystyle\int\frac{\csc^{4}x}{\cot^{2}x}dx=\int\frac{1}{\sin^{4}x}\cdot\frac{\sin^{2}x}{\cos^{2}x}dx=\int\frac{1}{\sin^{2}x\cos^{2}x}dx=\int\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}dx$

$\displaystyle=\int\left(\frac{1}{\cos^{2}x}+\frac{1}{\sin^{2}x}\right)dx=\int(\sec^{2}x+\csc^{2}x)dx=\tan x-\cot x+C.$

user84413
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show that $$\frac{\csc(x)^4}{\cot(x)^2}=\frac{1}{\sin(x)^2\cos(x)^2}$$ and use the tan half angle substitution.

Dr. Sonnhard Graubner
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