Integration Question

  • Thread starter Hyari
  • Start date
  • #1
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Homework Statement


integral cos^5(9*t) dt


Homework Equations


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?
 

Answers and Replies

  • #2
150
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Hint:
cos^2 u = 1-sin^2 u
 
  • #3
13
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u = cos(9t)
du = -9sin(9t)

[ 1 - sin(9t)^2 ] * [ 1 - sin(9t)^2 ] * u du?
 
  • #4
184
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[tex]\cos^5{9t}[/tex] is the same as [tex]\cos{9t}(1-\sin^2{9t})^2[/tex]

Expand it and integrate
 
  • #5
13
0
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 ?
 
  • #6
150
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try again, try u = something else.
 
  • #7
184
0
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 [tex]\frac{d}{dx}\sin{x} = \cos{x}[/tex]
 
  • #8
Gib Z
Homework Helper
3,346
5
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
13
0
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
Gib Z
Homework Helper
3,346
5
Do you know the Identity [itex]\sin^2 x + \cos^2 x=1[/itex]?
 
  • #11
HallsofIvy
Science Advisor
Homework Helper
41,833
962
u = cos(9t)
du = -9sin(9t)

[ 1 - sin(9t)^2 ] * [ 1 - sin(9t)^2 ] * u du?
What if you let u= sin(9t) instead?
 

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