# Math Help - Trigonometric integral cot^5(x)

1. ## Trigonometric integral cot^5(x)

I need to integrate:
$
\int \cot^{5}\left(3x\right)dx
$

I started to set u=3x then 1/3du=dx and I get

$
\frac{1}{3}\int \cot^{5}\left(u\right)dx
$

I know that
$
\frac{d}{dx} \cot x = -\csc^{2} x
$

and that

$
\cot^{2} x = \csc^{2} x -1
$

But now I am not sure how to go on. In the case of cot^3 I would write it as cotx and cot^2 x and set t=cotx.
Since I have cot^5 I do not know what to do.
Thanks for any help.

2. We have $\cot^5 u =\frac{\cos^5 u}{\sin ^5 u}
=\frac{\cos u \left(1-\sin^2 u\right)^2}{\sin ^5 u}$
so you can try to set $t =\sin u$

Is there a solution without writing it in sine-cosine?

I was looking for a solution with the identities with cot and csc.

And how I would go on if I change it to sine and cosine. I do not know how to do that.

Thanks

4. ## Re: Trigonometric integral cot^5(x)

To continue with the integral converted to sine and cosine, recall that the derivative of sin(x) is just cos(x)dx.
So what you want to do is multiply the polynomial in the numerator out. Do you then see how you will have a
set of terms containing some power of sin(x) times a cos(x)? All of these terms are still over a sin^5(x) of course.
You may now separate these terms into separate integrals.
Then you solve using u substitution with:
u = sin(x)
du = cos(x) dx

Good Luck!

5. ## Re: Trigonometric integral cot^5(x)

Originally Posted by DBA
I need to integrate:
$
\int \cot^{5}\left(3x\right)dx
$

I started to set u=3x then 1/3du=dx and I get

$
\frac{1}{3}\int \cot^{5}\left(u\right)dx
$

I know that
$
\frac{d}{dx} \cot x = -\csc^{2} x
$

and that

$
\cot^{2} x = \csc^{2} x -1
$

But now I am not sure how to go on. In the case of cot^3 I would write it as cotx and cot^2 x and set t=cotx.
Since I have cot^5 I do not know what to do.
Thanks for any help.
\displaystyle \begin{align*} \cot^5{(u)} &\equiv \cot^3{(u)}\cot^2{(u)} \\ &\equiv \cot^3{(u)} \left[ \csc^2{(u)} - 1 \right] \\ &\equiv \cot^3{(u)}\csc^2{(u)} - \cot^3{(u)} \\ &\equiv \cot^3{(u)}\csc^2{(u)} - \cot{(u)}\cot^2{(u)} \\ &\equiv \cot^3{(u)} \csc^2{(u)} - \cot{(u)}\left[ \csc^2{(u)} - 1 \right] \\ &\equiv \cot^3{(u)} \csc^2{(u)} - \cot{(u)}\csc^2{(u)} - \cot{(u)} \\ &\equiv \left[ \cot^3{(u)} - \cot{(u)} \right] \csc^2{(u)} - \cot{(u)} \end{align*}

So that means

\displaystyle \begin{align*} \frac{1}{3} \int{ \cot^5{(u)} \, du} &= \frac{1}{3} \left\{ \int{ \left[ \cot^3{(u)} - \cot{(u)} \right] \csc^2{(u)} \,du} - \int{ \cot{(u)}\,du} \right\} \end{align*}

You can solve the first integral by substituting \displaystyle \begin{align*} t = \cot{(u)} \end{align*} and the second you can look up from tables or set \displaystyle \begin{align*} \cot{(u)} = \frac{\cos{(u)}}{\sin{(u)}} \end{align*} and substituting \displaystyle \begin{align*} v = \sin{(u)} \end{align*}.

6. ## Re: Trigonometric integral cot^5(x)

I think the final answer works out to:

integral cot^5(x) dx = 1/4sin^4(x) - sin^2(x) + ln|sin(x)| + C

Do you concur?

7. ## Re: Trigonometric integral cot^5(x)

Newbie question: How do I get the fancy math formatting?

8. ## Re: Trigonometric integral cot^5(x)

I get $\int \cot^5x \,dx = -\frac{1}{4}\csc^4{x}+\csc^2{x}+\ln(|\sin{x}|)+C$

The "fancy math formatting" is called LaTeX. It's pretty easy to learn and there's a lot of information online about it. You start a formula with [TEX] and end with [/TEX]. If you mouse over a formula here on MHF, it gives you the LaTeX code used, which is really nice. There's also a LaTeX help forum here: LaTeX Help - Math Help Forum.

If you go to the Advanced editor, there's a $\Sigma$ that puts in the [TEX] and [/TEX] for you. You can also copy and paste formulas from previous posts.

- Hollywood