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Confirm this easy integral

  1. Apr 7, 2009 #1
    This is not a homework question. I am doing extra work to get a grasp on integration and I am not sure if this is correct...

    [tex]\int(sin^2x*cos^2x)dx[/tex]

    I did this:

    1) [tex]\int(sinx*cosx)*(sinx*cosx)dx[/tex]

    2) [tex]\int(1 /2 sin(2x)) * (1 /2 sin(2x)) dx[/tex] (Double angle for sine formula)

    3) [tex](1/4) \int(sin^2(2x))dx[/tex]

    4) [tex](1/4) * (1/2) \int(1-cos2x)dx[/tex] (Reduction of square power)

    5) = 1/8[x + (1/2)sin2x] + C

    I am not sure if this is correct, just hoping someone could help me out a little or confirm it ^^

    Cheers,
    Adrian
     
  2. jcsd
  3. Apr 7, 2009 #2
    Almost right, except that in going from step 3 to 4, you should have instead
    [tex](1/4) * (1/2) \int(1 - \cos 4x) \,dx.[/tex]
    You also have the sign on the sin term wrong in going from step 4 to 5 (the antiderivative of cos is sin, not -sin). The answer would then be
    [tex]\frac18 \left( x - \frac14 \sin 4x \right).[/tex]
     
  4. Apr 7, 2009 #3
    Thanks adriank, its good to know I was at least partly there ^^

    Cheers,
    Adrian
     
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