1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Indefinite Integral Homework

  1. May 3, 2008 #1
    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!
     
  2. jcsd
  3. May 3, 2008 #2
    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?
     
  4. May 3, 2008 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  5. May 4, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Why do you say trigonometric integrals have special rules? AFAIK they have the same rules as other integrals. What's special?
     
  6. May 4, 2008 #5
    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.
     
  7. May 4, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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: May 4, 2008
  8. May 4, 2008 #7
    Hint:

    You need to reduce the power of the sin from [tex] \sin{^3 (13x)} [/tex] into [tex] \sin{13x} [/tex]
     
  9. May 4, 2008 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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".
     
  10. May 4, 2008 #9
    I tried that formula and got it incorrect...
     
  11. May 4, 2008 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's not very helpful. How are we supposed to know how you got it wrong?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Indefinite Integral Homework
  1. Indefinite Integration (Replies: 11)

  2. Indefinite Integrals (Replies: 4)

  3. Indefinite Integral (Replies: 2)

Loading...