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!

Integral with multiple answers

  1. Aug 16, 2012 #1
    The question is ∫3cos25x dx. I tried to solve it 2 ways and got two different answers.

    Method A.
    ∫3cos25x dx =3∫cos25x dx

    u=5x
    3∫cos2u (1/5)du/dx dx
    The 2 dx cancel out leaving
    3∫cos2u (1/5)du=
    3/5 ∫cos2u du=
    3/5 ∫(1/2 + 1/2 cos 2u) du=
    3/5( (1/2x)+∫1/2 cos 2u du)=
    3/5( (1/2x)+1/2∫ cos 2u du)=
    3/5( (1/2x)+1/2(1/2sin2u))=
    3/5( (1/2x)+1/4sin2u)=
    3/10x + 3/20 sin 2u=
    3/10x + 3/20 sin 10x +C

    Method B
    ∫3cos25x dx =3∫cos25x dx=

    3∫(1/2 + 1/2cos 2(5x)) dx=
    3∫(1/2 + 1/2cos 10x) dx=
    3∫(1/2(1+cos10x)) dx=
    3/2 ∫(1+cos10x) dx=
    3/2 (∫1 dx+ ∫cos10x dx)=
    3/2(x+∫cos10x dx)=
    3/2x+(3/2)(1/10 sin 10x)=
    3/2x+ 3/20 sin10x +C

    Did I do something wrong in either of the methods? thanks!
     
    Last edited: Aug 16, 2012
  2. jcsd
  3. Aug 16, 2012 #2

    eumyang

    User Avatar
    Homework Helper

    Here, it should be [itex]\frac{1}{2}u[/itex]. Then substitute the 5x in for u.
     
  4. Aug 16, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In the first method you took the integral of the constant 1/2 as (1/2)x instead of the required (1/2)u. You are integrating wrt u, not x.

    RGV
     
  5. Aug 16, 2012 #4
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integral with multiple answers
  1. Multiple integrals (Replies: 6)

  2. Multiple Integrals (Replies: 2)

  3. Multiple Integral (Replies: 6)

  4. Multiple integrals (Replies: 6)

Loading...