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Homework Help: 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

    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


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    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

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    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.

  5. Aug 16, 2012 #4
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