# Integral cos^5(9t) dt: Solving Path

• Hyari
u = sin(9t)du = 9cos(9t) dt[ 1 - u^2 ] * [ 1 - u^2 ] * (1/9)du(1/9) * [ 1 - u^2 ]^2 * (1-u^2)du(1/9) * [ (1 - 2u^2 + u^4) * (1 - u^2) ]du(1/9) * [ 1 - 3u^2 + 3u^4 - u^6 ]du(1/9) * [ u - u^3 + u^5 - (1/3)

## Homework Statement

integral cos^5(9*t) dt

half sets?

## 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?

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

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

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

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

Expand it and integrate

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 ?

try again, try u = something else.

Hyari said:
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 ?

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

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.

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 :(

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

Hyari said:
u = cos(9t)
du = -9sin(9t)

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

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