# Integration Question

1. Feb 25, 2007

### Hyari

1. The problem statement, all variables and given/known data
integral cos^5(9*t) dt

2. Relevant equations
half sets?

3. The attempt at a solution
integral cos(9t)^5 dt

integral cos(9t)^2 * cos(9t)^2 * cos(9t)

cos(9t)^2 = (1/2)*[ 1 + cos(18t) ]

integral (1/2)*[ 1 + cos(18t) ] * (1/2)*[ 1 + cos(18t) ] * cos(9t)

Am I on the right path?

2. Feb 25, 2007

### gammamcc

Hint:
cos^2 u = 1-sin^2 u

3. Feb 25, 2007

### Hyari

u = cos(9t)
du = -9sin(9t)

[ 1 - sin(9t)^2 ] * [ 1 - sin(9t)^2 ] * u du?

4. Feb 26, 2007

### theperthvan

$$\cos^5{9t}$$ is the same as $$\cos{9t}(1-\sin^2{9t})^2$$

Expand it and integrate

5. Feb 26, 2007

### Hyari

Then how do you integrate that beat .

cos(9t) * (1 - sin(9t)^2)^2 * dt

u = 1-sin(9t)^2
du = -18sin(9t) * cos(9t) * dt

du / -18sin(9t) = cos(9t) * dt

1 / -18sin(9t) <-integral-> u^2 * du

1/-18sin(9t) * u^3

1/-18sin(9t) * (1-sin(9t)^2)^3 ?

6. Feb 26, 2007

### gammamcc

try again, try u = something else.

7. Feb 26, 2007

### theperthvan

You need to expand it. Then you will be able to use $$\frac{d}{dx}\sin{x} = \cos{x}$$

8. Feb 26, 2007

### Gib Z

So basically when we have to integrate something with only cosine in it, and cosine is odd powered, we take the most even powers out and transform those into the sines as mentioned, expand and use substitution u=sin x to do the rest.

9. Feb 26, 2007

### Hyari

I don't understand... can you give me an example?

cos^2 * cos^2 * cos(x) = (1 - sin^2)^2 * cos(x).

I don't understand :(

10. Feb 27, 2007

### Gib Z

Do you know the Identity $\sin^2 x + \cos^2 x=1$?

11. Feb 27, 2007

### HallsofIvy

Staff Emeritus
What if you let u= sin(9t) instead?