ANGAMWhat is the Integral of Cube Root of Cot(x)?

[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.

Homework Statement

f(x) = ³\sqrtcotx

Find : \int f(x)

Indefinite Integral

Homework Equations

[Basic Integration Laws]

The Attempt at a Solution

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

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Looking forward to your reply
Shubhangam

 

[Hurkyl]

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?

 

[shubhangam]

First, thanks for your fast reply

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...

 

[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.

 

[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)

 

[Hurkyl]

First, thanks for your fast reply

I substituted cot^3(x) as z

getting \int z dx

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?

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...

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)

 

[shubhangam]

Okay I'm getting

<br /> <br /> \int \frac {3 z^3 \, dz}{1 + z^6}<br /> <br />

Using z=³\sqrtcot(x)

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

 

[Hurkyl]

Factorize the denominator and use partial fractions?

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)

 

[shubhangam]

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)

So.. any faster method?

 

[arildno]

Okay I'm getting

<br /> <br /> \int \frac {3 z^3 \, dz}{1 + z^6}<br /> <br />

Using z=³\sqrtcot(x)

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

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

 

[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