# Odd power of sine Integral

1. May 6, 2007

### tanky322

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

Integral((sin(x))^2((cos(x))^2) dx

2. Relevant equations

3. The attempt at a solution

Seperate Cos (x)^3

sin(x)^2 (cos(x))(cos(x))^3

Then:apply identities

sin(x)^2(cos(x))(1-sin(x)^2)

And now im lost!!!!

Thanks Alot!!!

2. May 6, 2007

### Mindscrape

Is that supposed to be (cosx)^4 or something (unless you divide by cos as well, you cannot have a cos^3 out of cos^2)?

The way I would do this is to use the identity that sin^2 x is 1 - cos^2 x, and then you can integrate cos^n x with a reduction formula or double angle identities.

Or you could start off with double angle identities and see where it leads you.

3. May 6, 2007

### tanky322

Yea Sorry I screwed that up the original integral is
Integral((sin(x))^2((cos(x))^3) dx

Then
Seperate Cos (x)^3

sin(x)^2 (cos(x))(cos(x))^2

Then:apply identities

sin(x)^2(cos(x))(1-sin(x)^2)

So now If I change the sin(x)^2 to 1-cos(x)^2 I wont have dx just as if I change it back????

4. May 6, 2007

### Noober

(sinx)^2 = 1-(cosx)^2
so...

Int((sin(x))^2((cos(x))^3) dx
=Int[(cosx)^3(1-(cosx)^2)]dx
=Int[(cosx)^3 - (cosx)^5]dx
= -(cosx)^4/4sinx + (cosx)^6/6sinx

Right?

Last edited: May 6, 2007
5. May 6, 2007

### Office_Shredder

Staff Emeritus
I would stick with changing cosine to sin. Then you get
$$\int sin^2(x)cos(x)dx -\int sin^4(x)cos(x)dx$$

A discerning eye will note a great substitution there

6. May 6, 2007

### IMDerek

Perhaps you should try using

$\sin (2x) = 2 \sin (x) \cos (x)$

edit: sorry, didn't see the new problem

7. May 7, 2007

### tanky322

Ok heres my shot at this one at 6:40 AM

so it becomes

$$\int sin^2(x)cos(x)dx -\int sin^4(x)cos(x)dx$$

by using 1-sin(x)^2

So I make u sin(x)
du cos(x)

Therefore I get (sin(x)^3)/3-(sin(x)^5)/5