# How do you take integral of cot ?

1. Mar 24, 2005

### EasyStyle4747

integral of (cot2x)^3 dx

are you supposet to convert to cos and sin?

2. Mar 24, 2005

### Data

Try using the identity $\csc^2{x} - \cot^2{x} = 1$ and later making the substitution $u = \csc{2x}$ (similar to integrals of powers of sines and cosines, or secants and tangents. Of course, you'll have to deduce the differential of u! ).

Last edited: Mar 24, 2005
3. Mar 24, 2005

### xanthym

The answer to your question for this case: YES (for easy solution)
Then sub {w = sin(2*x) ::AND:: dw = 2*cos(2*x)*dx ::OR:: (dw/2) = cos(2*x)*dx}:

∫ {cot(2*x)}^3 dx = ∫ {cos(2*x)/sin(2*x)}^3 dx =
= ∫ {cos(2*x)}^2/{sin(2*x)}^3 {cos(2*x)}dx =

= ∫ (1 - w^2)/w^3 {dw/2} = ::: Sub "w" using {cos^2() = 1 - sin^2()} in num
= (1/2)*∫ {w^(-3) - w^(-1)} dw =
::: (integrating) :::

= (1/2)*{(-1/2)*w^(-2) - Loge(|w|)} + C =
= (-1/4)*{sin(2*x)}^(-2) - (1/2)*Loge{|sin(2*x)|} + C =

= (-1/4)*{csc(2*x)}^2 - (1/4)*Loge{{sin(2*x)}^2} + C

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Last edited: Mar 25, 2005
4. Mar 24, 2005

### Data

Here's the solution using my substitution:

Note that $u = \csc(2x)$ and thus $du = -2\csc(2x)\cot(2x)dx$

$$\int \cot^3(2x) \ dx = \int (\csc^2(2x) - 1)\cot(2x) \ dx$$
$$= \frac{-1}{2}\int \frac{u^2 - 1}{u} \ du = \frac{-1}{2}\left( \frac{u^2}{2} - \ln{|u|} \right) + C$$

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

Last edited: Mar 24, 2005
5. Mar 24, 2005

### EasyStyle4747

wow, 2 different solutions! thanks a lot guys, this helps a lot.

6. Mar 24, 2005

### xanthym

The 2 solutions presented in this thread are equivalent in final results. Note that:
csc^2() = sin^(-2)()
::: ⇒ Loge{csc^2()} = (-1)*Loge{sin^2()}

Thus, we have:

$$:(1): \ \ \ \ \frac{1}{4}\ln \left(\csc^2(2x)\right) \ - \ \frac{1}{4}\csc^2(2x) \ + \ C \ \ = \ \ \frac{-1}{4}\csc^2(2x) \ + \ \color{red} (-1)*\color{black}\frac{1}{4}\ln \left(sin^2(2x)\right) \ + \ C$$

which is equivalent to that presented in the EASY ( ) solution in msg #3 involving only sin() & cos() functions, identities, & derivatives (except for the very last equation)!!

~~

Last edited: Mar 25, 2005
7. Mar 24, 2005

### Data

There's no error. I just squared the csc inside the logarithm and as a result pulled out an extra $$\frac{1}{2}$$. Doing it that way gets rid of the absolute value signs for one thing. Note that in your original solution the argument to $\ln$ was $\sin{2x}$, not $\sin^2{2x}$

8. Mar 24, 2005

### xanthym

Indeed, there was no error. And the squared Loge argument is a good idea. After adjusting orig msgs, both solutions are now equiv in final result.

~~

Last edited: Mar 25, 2005
9. Mar 24, 2005

### Data

Well, we actually did both make a mistake. Pesky constants of integration