1. Limited time only! Sign up for a free 30min personal 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!

Calculus 2 Integral Question (difficult)

  1. Sep 24, 2008 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    cos3x = cos2x * cosx

    3. The attempt at a solution

    Let u= cos2x
    du=(x/2) + (cos2x/4x)


    -cos2xsinx - [tex]\int[/tex]-sinx(x/2 + cos2x/4x)dx

    -cos2xsinx -[tex]\int[/tex]xsinx/2 + (sinxcos2x/4x)

    -cos2xsinx - (1/2)xsinx +(sinxcos2x/4x)

    -cos2xsinx -(sinx-xcosx/2) + (sinxcos2x/4x)


    aaaand that's about where I get lost/stuck. i can't figure out how to finish off the problem. i get to the last line on the far right and get confused. any help would be much appreciated.
  2. jcsd
  3. Sep 24, 2008 #2

    Why don't you try and use the trig identity[tex] \cos^2x +\sin^2x = 1 [/tex] and then try substituting [tex] \sin x = u [/tex]

    Hope that helps,

  4. Sep 24, 2008 #3
    hm ill give it a shot
  5. Sep 24, 2008 #4
    No worries, glad to have helped.

    - Spoon
  6. Sep 24, 2008 #5


    User Avatar
    Science Advisor

    Here's your first error. The derivative of cos2 x, using the chain rule, is (2 cos x)(cos x)'= - 2sin x cos x. I have absolutely no idea how you got your du!

    You started wrong by trying to use integration by parts when it is not necessary. cos3 x dx= (cos2 x) (cos x dx)= (1- sin2 x) (cos x dx). Now use the obvious substitution.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Calculus 2 Integral Question (difficult)