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Basic Integration win sin and cos

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Integrate [tex]\int\sin 6x + 3\cos 5x dx[/tex]

    2. Relevant equations

    3. The attempt at a solution

    The way I was taught was the 'u' substitution method.
    I know that [tex]\int\sin u du = -\cos u + C[/tex] and I know that [tex]\int\cos u du = \sin u + C[/tex]

    Here's what I've done so far

    [tex]\int\sin 6x + 3\cos 5x dx[/tex]
    [tex]\int\sin 6x dx + 3\int\cos 5x dx[/tex]
    u = 6x; u = 5x
    du = 6dx; du = 5dx
    du/6 = dx; du/5 = dx

    [tex]\frac{1}{6}\int\sin u du + \frac{3}{5}\int\cos u du[/tex]

    Am I on the right track?
    Last edited: Apr 28, 2008
  2. jcsd
  3. Apr 28, 2008 #2
    Yep, everything you've done looks good so far. Just evaluate the integrals and sub the u values back in and you should be finished.
  4. Apr 28, 2008 #3
    Answer: [tex]\frac{-1}{6}\cos 6x + \frac{3}{5}\sin 6x + C[/tex] ?
  5. Apr 28, 2008 #4


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    Your're OK so far...

    Such is the glacial pace of my ancient home computer... Yes, your antiderivative is correct.
    Last edited: Apr 28, 2008
  6. Apr 28, 2008 #5
    Close, not quite. Keep in mind that you have two separate u values for the two different integrals.

    (Sometimes it's easier to use both u's and v's to avoid confusion)
  7. Apr 28, 2008 #6
    \frac{-1}{6}\cos 6x + \frac{3}{5}\sin 5x + C
  8. Apr 28, 2008 #7
    Yep, that's the answer I got. It seems like you've got the calculus down!

    Sometimes the algebra is all that will mess you up, lol.
  9. Apr 28, 2008 #8
    Yeah, I don't have as strong of an Algebra background as I should.
    Thanks for the help
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