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Math Help - Problems with a few Integrals

  1. #1
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    Problems with a few Integrals

    I just started Calc II and the first thing the teacher did was give us a review sheet of about 60 integrals and derivatives. There are 3 integrals in particular that I have been going around in circles with.

    $=Integral

    1) $cos^2[(1/2)x]dx

    I know that the answer wanted is (1/2)x + (1/2)sinx + C and that the half angle formula gets me on on the correct path but I am missing something...

    2) $sin^2(3x)dx

    Again, I have been given the answer (1/2)x - (1/12)sin(6x) + C. This gives a hint to use the double angle forumua but I'll be darned if I can make it work.

    3) $sin^4(x)dx

    The answer given is (1/32)[12x-8sin(2x) + sin(4x)] + C. I have tried breaking this into (sin^2(x))^2 = (1 - cos^2(x))^2 = 1 - 2cos^2(x) + cos^4(x) and again I am doing something wrong.

    I am sorry for posting 3 at a time. Any help/hints will be greatly appreciated!

    Thank you in advance,

    Quasar
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  2. #2
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    Quote Originally Posted by Quasar View Post

    1) $cos^2[(1/2)x]dx
    Use half-angle identity,
    \int \frac{1}{2}+\frac{1}{2}\cos x dx
    The rest is simple,
    \frac{1}{2}x+\frac{1}{2}\sin x+C
    2) $sin^2(3x)dx
    Again half-angle identity,
    \int \frac{1}{2}-\frac{1}{2}\cos 6x dx
    Now,
    \int \frac{1}{2} dx=\frac{1}{2}x
    And,
    \int \frac{1}{2} \cos 6x dx = \frac{1}{12}\sin 6x
    If you use u=6x.
    Thus,
    \frac{1}{2}x-\frac{1}{12}\sin 6x+C
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  3. #3
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    Quote Originally Posted by Quasar View Post
    3) $sin^4(x)dx
    You have,
    \int \sin^4 x dx
    \int (\sin^2 x)^2 dx
    Use half-angle identity,
    \int \left( \frac{1}{2}-\frac{1}{2}\cos 2x \right)^2 dx
    Open sesame,
    \int \frac{1}{4}-\frac{1}{2}\cos 2x+\frac{1}{4}\cos^2 2x dx
    Use half-angle identity again,
    \int \frac{1}{4}-\frac{1}{2}\cos 2x +\frac{1}{4}\left( \frac{1}{2}+\frac{1}{2}\cos 4x \right) dx
    Open sesame,
    \int \frac{1}{4}-\frac{1}{2}\cos 2x +\frac{1}{8}+\frac{1}{8}\cos 4x dx
    Combine the brethren,
    \int \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x dx
    Thus,
    \frac{3}{8}x-\frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x +C
    Factor,
    \frac{1}{32} \left( 12x - 8\sin 2x+\sin 4x\right)+C
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