# Integrate cosx^3dx

1. Jan 24, 2004

### lastlaugh

Any suggestions on how to integrate
$$\int cos^3xdx$$
My first clue would be to substitute
(1-(Sinx)^2)cosxdx but any clue to go after this?

Last edited: Jan 24, 2004
2. Jan 24, 2004

### Hurkyl

Staff Emeritus
Well, do any of the techniques you've learned apply to

$$\int (1 - \sin^2 x) \cos x \, dx$$

?

3. Jan 24, 2004

### himanshu121

Put sinx=t => cosx dx=dt

so u have int(1-t2)dt=t-t3/3

4. Jan 24, 2004

### lastlaugh

thanks for the help. This trig integration stuff is killer

OK the reason I asked was because I have a lab assignment where I have to calculate the powers of $$\int cos^n[x]dx$$ where n is odd numbers from 3-13.
So can someone please look over my work for $$\int cos^5[x]dx$$ and make sure it looks valid:
First I substituted $$(1-sin^2[x])^2$$ for $$cos^4[x]$$ which gives me:
$$\int (1-sin^2[x])^2 cos[x]dx$$
Then I let u=sinx meaning du=cosxdx
Giving me:
$$\int (1-u^2)^2du$$
Which I simplfied to give me:
$$\int u^4 du - \int 2u^2 du + \int du$$
Which gives me
$$(u^5)/5-(2u^3)/3 + u$$

Anyone see any glaring errors?

Last edited: Jan 24, 2004
5. Jan 24, 2004

### PrudensOptimus

WRONGGGGGGGGGGGGGGGGGGGG

is (1-u^2) du.

6. Jan 25, 2004

### Hurkyl

Staff Emeritus
Looks right to me. (except you should substitute back what u is)

7. Jan 25, 2004

### himanshu121

How it is wrong

Moreover General way will be to form a reduction formula

Last edited: Jan 25, 2004
8. Jan 25, 2004

### lastlaugh

Yeah I didn't feel like typing out any more latex