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Math Help - trig powers integration

  1. #1
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    trig powers integration

    \int (8x+sinx)^2dx

     \int (sinx^2+16xsinx+64x^2)dx

     \int (\frac{1-cos2x}{2}+16xsinx+64x^2)dx

    this is where I get stuck
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  2. #2
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    Quote Originally Posted by silencecloak View Post
    \int (8x+sinx)^2dx

     \int (sinx^2+16xsinx+64x^2)dx

     \int (\frac{1-cos2x}{2}+16xsinx+64x^2)dx

    this is where I get stuck
    You can split your last sum to get

    \int \frac{1-cos(2x)}{2}dx + \int 16xsin(x) dx + \int 64x^2 dx

    <br />
\int \frac{1-cos(2x)}{2}dx = \frac{1}{2}\int -\frac{cos(2x)}{2} = -\frac{1}{4}sin(2x) (eq1)

    the next one needs the substitution rule:

    u = 16x ; u' = 16
    v' = sin(x) ; v = cos(x)

    \int uv^{\prime} = uv - \int vu^{\prime}

    16xcos(x) - \int (16cos(x))dx = 16(x - sin(x)) (eq2)

    the last one is simple

    \int 64x^2 = \frac{64}{3}x^3 (eq3)

    Add our equations together (not forgetting the constant of integration):

    [-\frac{1}{4}sin(2x)] + [16(x - sin(x))] + \frac{64}{3}x^3 + C
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  3. #3
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    Ive never seen some of those techniques...
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  4. #4
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    I'm not sure that is correct.

    Webassign (online math) won't take that, but I could be wrong

    Either way

    Could someone please help me through this with a more generic approach?
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