Indefinite Integral Homework

  • Thread starter MillerL7
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  • #1
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The integral of [sin^3*(13x)*cos^8*(13x)]dx

I think u=sin^3
so du/dx=cos^3
du=cos^3dx

but then I am really not sure if that is correct, the trig functions confuse me a bit. Please help me, thank you!
 

Answers and Replies

  • #2
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I don't think that you can simply use a u-substitution here, this appears to be a Trigonometric Integral to me, which has its own special rules.

Since this is in the form of [itex] Sin^n(u)Cos^m(u) [/itex] where m and n are both integers. Where does your book tell you to start?

Hint: focus on m and n, what does that tell you?
 
  • #3
Dick
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The chain rule also seems to be confusing you. If u=sin^3(x) then du=3*sin^2(x)*cos(x), NOT cos^3(x). Please review that before even attempting this problem. And the fact that sin^2(x)=1-cos^2(x) may help you a lot. If you play your cards right, it's a pretty simple integral using u substitution.
 
  • #4
Dick
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I don't think that you can simply use a u-substitution here, this appears to be a Trigonometric Integral to me, which has its own special rules.

Since this is in the form of [itex] Sin^n(u)Cos^m(u) [/itex] where m and n are both integers. Where does your book tell you to start?

Hint: focus on m and n, what does that tell you?
Why do you say trigonometric integrals have special rules? AFAIK they have the same rules as other integrals. What's special?
 
  • #5
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I guess I phrased that poorly, I simply meant that the integral was in a specific form which can be dealt with easily when you look at the powers of Sin and Cos. In this case he would want to break up [itex] Sin^3(13x) [/itex] using [itex] Sin^2(x) + Cos^2(x) = 1 [/itex] because the power of the sine function is odd.
 
  • #6
Dick
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I guess I phrased that poorly, I simply meant that the integral was in a specific form which can be dealt with easily when you look at the powers of Sin and Cos. In this case he would want to break up [itex] Sin^3(13x) [/itex] using [itex] Sin^2(x) + Cos^2(x) = 1 [/itex] because the power of the sine function is odd.
Right. Sorry, I just jumped on you when you said you can't use a u substitution. Because the answer IS a u substitution. Together with a specific trig rule.
 
Last edited:
  • #7
Hint:

You need to reduce the power of the sin from [tex] \sin{^3 (13x)} [/tex] into [tex] \sin{13x} [/tex]
 
  • #8
HallsofIvy
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In general if you have an odd power of a sine or cosine you can factor one of them out to use with the dx, then reduce the remaining even power by sin2x= 1- cos2 x or cos2x= 1- sin2x.

Here you have [sin3(13x) cos8(13x)]dx which has sine to an odd power. Write it as [sin2(13x) cos8(13x)] sin(x)dx= (1- cos2(13x))cos8(13x)] sin(13x) dx and let u= sin(13x). Be careful with the "13x".
 
  • #9
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I tried that formula and got it incorrect...
 
  • #10
Dick
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I tried that formula and got it incorrect...
That's not very helpful. How are we supposed to know how you got it wrong?
 

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