# Homework Help: Integral Cube Root of Cot(x)

1. Jul 5, 2008

### shubhangam

Hi, I just registered on to the forums, and I have to say it is a very good job. I am currently a senior at High School and aiming for an Engineering Course.

While solving some integrals, I got stuck upon this one, and even after a lot of attempt, my friends and I could not solve it. Hence I need your help.

1. The problem statement, all variables and given/known data

f(x) = 3$$\sqrt$$cotx

Find : $$\int$$ f(x)

Indefinite Integral

2. Relevant equations

[Basic Integration Laws]

3. The attempt at a solution

A whole lot of pages.. I have no idea how much to put in here :|

---
Shubhangam

2. Jul 5, 2008

### Hurkyl

Staff Emeritus
Well, seeing how you really only know how to antidifferentiate (into a 'closed form') very few expressions involving cube roots, that suggests two obvious substitutions to try. Do you see what they are? Where did you get stuck when trying them?

3. Jul 6, 2008

### shubhangam

I substituted cot^3(x) as z

getting $$\int$$ z dx

Now dz/dx = -3 cot^2 (x) cosec^2 (x)
so, dx = dz/[-3 cot^2 (x) cosec^2 (x)]

I replaced that in the integral

I have no idea how to proceed...

4. Jul 6, 2008

### ace123

Well first you substituted the wrong thing. Your problem is $$\sqrt[3]{cot x}$$. Yet your subsitution was cot^3(x). Care to explain that. Which is your actual problem. Also I think their is an easier substitution. Clarify your problem.

5. Jul 6, 2008

### arildno

Let us do this properly:
$$z(x)=\cot(x)^{\frac{1}{3}}$$
$$\frac{dz}{dx}=\frac{1}{3}\cot^{-\frac{2}{3}}*(-\frac{1}{\sin^{2}(x)})=-\frac{1}{3}z^{-2}(1+\cot^{2}(x))=-\frac{1}{3}(z^{-2}+z^{4})$$

It ain't pretty, but you can solve it..

You might have use for the factorization:
$$x^{6}+1=(x^{2}-\sqrt{3}x+1)(x^{2}+1)(x^{2}+\sqrt{3}x+1)$$

Last edited: Jul 6, 2008
6. Jul 6, 2008

### Hurkyl

Staff Emeritus
Well, that's your first problem. The relation you used to make the substitution was

$$z = \cot^3 x$$

but you asserted that the integrand satisfies

$$z = \cot^{1/3} x$$

which contradicts the fact $\cot^{1/3} x \neq \cot^3 x$!

I suppose you meant to do either $z = \cot^{1/3} x$, or maybe simply $z = \cot x$?

Keeping in mind that this particular form is wrong (by the previous reason) -- you're saying that you got

$$\int \frac{z \, dz}{-3 \cot^2(x) \csc^2 x}$$

right? So is the problem that you don't know what you need to do, or simply that you don't remember enough trigonometry to do it? (or that you simply didn't have enough confidence to continue working?)

(Even though this is wrong, you will be faced with a similar situation when you get the substitution right)

7. Jul 6, 2008

### shubhangam

Okay I'm getting

$$\int \frac {3 z^3 \, dz}{1 + z^6}$$

Using z=3$$\sqrt$$cot(x)

But, how do I proceed after that? Factorize the denominator and use partial fractions? or is there some part I am missing?

8. Jul 6, 2008

### Hurkyl

Staff Emeritus
That looks similar to something I got. Anyways, partial fractions always works for rational functions. (although, the roots, and thus the factors, aren't always simple expressions)

9. Jul 6, 2008

### shubhangam

So.. any faster method?

10. Jul 6, 2008

### arildno

I've already provided you with that factorization, dear.

11. Jul 6, 2008

### shubhangam

Hmm, its getting late atm, I will work on this tomorrow and get back

Thanks for the help

Also, just for reference, is there a better method for this integral, this substitution method seems to be pretty long.. Or is this the only method?

sHubH