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Calculus help

by SEG9585
Tags: calculus
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SEG9585
#1
Mar18-04, 03:41 PM
P: 34
Hey all--
I had an Integration of Parts quiz today and got stuck on a few problems-- was wondering if you could explain the steps involved in solving these integrals:

int( (sin(3x))^3 * (cos(3x))^3 dx)

and

int( (tan(4x))^4) dx)

thanks!!
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cookiemonster
#2
Mar18-04, 04:05 PM
P: 991
[tex]\int \sin^3{3x} \cos^3{3x}\,dx[/tex]

First things first, get rid of the 3x's with a u substitution. It's easy to see that all it does is change the solution by a factor of 1 over 3. So we have,

[tex]\frac{1}{3}\int \sin^3{u} \cos^3{u}\,du[/tex]

If we had only 1 sine term or only 1 cosine term, we'd be gold. Problem solved. But we got 2 too many. So let's get rid of them!

[tex]\sin^2{x} + \cos^2{x} = 1[/tex]

Use this to turn the integral into

[tex]\frac{1}{3}\int \sin^3{u} (1 - \sin^2{u}) \cos{u}\,du[/tex]

which is easily separated and solved by substitution.

Have another shot at the second one, keeping in mind that

[tex]\frac{d}{dx}\tan{x} = \sec^2{x}[/tex]

cookiemonster
NateTG
#3
Mar18-04, 04:52 PM
Sci Advisor
HW Helper
P: 2,538
Where did you get [tex]\cos^3=1-\sin^2[/tex]
I would probably use the half angle formulas:
[tex]\sin(2x)=2\sin(x)\cos(x)[/tex]
so
[tex]\sin(x)\cos(x)=\frac{\sin(2x)}{2}[/tex]
so
[tex]\sin(3x)\cos(3x)=\frac{\sin(6x)}{2}[/tex]

Now you've got:
[tex]\int\sin^3(3x)\cos^3(3x)dx[/tex]
[tex]\int(\sin(3x)\cos(3x))^3dx[/tex]
[tex]\frac{1}{8}\int\sin^3(6x)dx[/tex]

Now
[tex]\sin(3x)=3\sin(x)-4\sin^3(x)[/tex]
so
[tex]\sin^3(6x)=\frac{3\sin(6x)-\sin(18x)}{4}[/tex]
so
[tex]\frac{1}{32}\int3\sin(6x)-\sin(18x)dx[/tex]
so
[tex]\int\sin^3(3x)\cos^3(3x)dx=\frac{\cos(18x)}{576}-\frac{\cos(6x)}{192}+C[/tex]

Oh, by parts...

cookiemonster
#4
Mar18-04, 04:55 PM
P: 991
Calculus help

I didn't. I only took two of the cosines and I left the third for the u substitution.

u = sinx
du = cosxdx

It's used in the du.

Edit: By parts?

cookiemonster


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