How do I solve the integral of cot^3(x)?

[november1992]

Homework Statement

Integrate
∫$cot^{3}(x)$

Homework Equations

u*v-∫vdu

The Attempt at a Solution

I used the integration by parts formula and I got:

$cot^{3}(x)$(ln|sin(x)|)-∫ln|sin(x)|-x-cot(x)dx

I don't know how to integrate the integrand.

 

[SammyS]

Homework Statement

Integrate
∫$cot^{3}(x)$

Homework Equations

u*v-∫vdu

The Attempt at a Solution

I used the integration by parts formula and I got:

$cot^{3}(x)$(ln|sin(x)|)-∫ln|sin(x)|-x-cot(x)dx

I don't know how to integrate the integrand.

Please ... let us know what you used for u & v. We could guess, but why make us guess ?

 

[LCKurtz]

Try writing $\cot^3(x) = \cot^2x\cot x = (\csc^2(x)-1)\cot x$. Then a couple of appropriate u-substitutions might work for you.

 

[november1992]

Please ... let us know what you used for u & v. We could guess, but why make us guess ?

$cot^{2}x$ as u, $-csc^{2}x$ as du
cot(x) as dv, ln|sin(x)| as v

Try writing $\cot^3(x) = \cot^2x\cot x = (\csc^2(x)-1)\cot x$. Then a couple of appropriate u-substitutions might work for you.

That was what I tried to do first. I made u=cot(x), du= -$csc^{2}x$
when i integrate i get $u^{3}$/3 - u

 

[LCKurtz]

Try writing $\cot^3(x) = \cot^2x\cot x = (\csc^2(x)-1)\cot x$. Then a couple of appropriate u-substitutions might work for you.

That was what I tried to do first. I made u=cot(x), du= -$csc^{2}x$
when i integrate i get $u^{3}$/3 - u

The u sub only works for the first term. Try writing $\cot x = \frac{\cos x}{\sin x}$ for the second term.

 

[november1992]

Okay, now i got:

∫(1-csc(x))*$\frac{cos(x)}{sin(x)}$

∫(1-$\frac{1}{sin(x)}$ * $\frac{cos(x)}{sin(x)}$

u=sin(x), du=cos(x)

∫(1-$\frac{1}{u}$)

u-lnu

sin(x)-ln|sin(x)|

 

[Dick]

Okay, now i got:

∫(1-csc(x))*$\frac{cos(x)}{sin(x)}$

∫(1-$\frac{1}{sin(x)}$ * $\frac{cos(x)}{sin(x)}$

u=sin(x), du=cos(x)

∫(1-$\frac{1}{u}$)

u-lnu

sin(x)-ln|sin(x)|

No, you don't have it yet. But you are getting closer. Write cot(x)^3=cos(x)^3/sin(x)^3. Now try u=sin(x).

 

[november1992]

I just ended up with ln|sin(x)| I don't know what I'm doing wrong.

 

[Dick]

I just ended up with ln|sin(x)| I don't know what I'm doing wrong.

It would be really hard to say what you are doing wrong if you don't show what you are doing. Now wouldn't it??

 

[november1992]

∫ cos(x)^3/sin(x)^3.

u=sin(x), du=cos(x)

∫1/u

ln(u)

ln|sin(x)|

 

[Dick]

∫ cos(x)^3/sin(x)^3.

u=sin(x), du=cos(x)

∫1/u

ln(u)

ln|sin(x)|

No. What happened the power 3? That would be ok, if it was cos(x)/sin(x). It's not. It's cos(x)^3/sin(x)^3.

 

[november1992]

∫$\frac{1}{u^3}$

$\frac{2}{sin^2}$

Is this right?

 

[Dick]

∫$\frac{1}{u^3}$

$\frac{2}{sin^2}$

Is this right?

Not even a little. If you substitute u=sin(x) du=cos(x) dx into $\int \frac{cos^3(x)}{sin^3(x)} dx$ you get $\int \frac{cos^2(x)}{u^3} du$. Now you just need to express cos(x)^2 in terms of u.

 

[november1992]

∫$\frac{u^2}{u^3}$

$\frac{2u^3}{u^2}$

$\frac{2cos^3}{cos^2}$



I don't think I'm doing it right. Trigonometric Integrals confuse me.

 

[Dick]

∫$\frac{u^2}{u^3}$

$\frac{u^3}{u^2}$

$\frac{cos^3}{cos^2}$. I don't think I'm doing it right. Trigonometric Integrals confuse me.

Yes, they do confuse you. cos(x)^2=1-sin(x)^2. Express that in terms of u.

 

[november1992]

Yes, they do confuse you. cos(x)^2=1-sin(x)^2. Express that in terms of u.

$\frac{1-sin^2(x)}{sin^3}$

 

[Dick]

$\frac{1-sin^2(x)}{sin^3}$

This is not going well. cos(x)^2=1-sin(x)^2=1-u^2. You might be too tired right now to think straight. I know I am. Gotta go now.

 

[Sefrez]

$\frac{1-sin^2(x)}{sin^3}$

That is not in terms of u. :p
u = sin x, so:
$∫\frac{1-u^2}{u^3}du$

 

[november1992]

This is not going well. cos(x)^2=1-sin(x)^2=1-u^2. You might be too tired right now to think straight. I know I am. Gotta go now.

I think it may be the fact that I'm very bad at math. Thanks for the help though, I appreciate it.

That is not in terms of u. :p
u = sin x, so:
$∫\frac{1-u^2}{u^3}du$

do i integrate that?

 

[Sefrez]

Also, just another way to solve, if you want some practice :

Mod note: Removed complete solution.[/color]

I think it may be the fact that I'm very bad at math. Thanks for the help though, I appreciate it.



do i integrate that?

Yes, split the faction into partial fractions. All you have is 1/u^3 - u^2/u^3.

 

[november1992]

Thanks for the help. I'll just read my textbook again and do some more practice problems.