Integration Question

[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?

[gammamcc]

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

[Hyari]

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

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

[theperthvan]

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

Expand it and integrate

[Hyari]

Then how do you integrate that beat o_O.

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 ?

[gammamcc]

try again, try u = something else.

[theperthvan]

Then how do you integrate that beat o_O.

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 ?

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

[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.

[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 :(

[Gib Z]

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

[HallsofIvy]

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

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

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