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Math Help - integral

  1. #1
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    Question integral

    -1/2 times the integral (from 0 to pi/2) of cos^4 theta - 2cos^2 theta d(theta)

    Thank you very much.
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  2. #2
    Senior Member tukeywilliams's Avatar
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     -\frac{1}{2} \int_{0}^{\pi/2} \cos^{4} \theta - 2 \cos^{2} \theta \ d \theta

    Use power reduction formulas:  \sin^{2} \theta = \frac{1- \cos 2 \theta}{2}

    and  \cos^{2} \theta = \frac{1+ \cos 2 \theta}{2}
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  3. #3
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    hey turkeywilliam,

    I did. But I just can't get the right answer. I got (5pi)/64 +1/8 . But the correct answer is (5pi)/32. Don't know why?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tukeywilliams View Post
     -\frac{1}{2} \int_{0}^{\pi/2} \cos^{4} \theta - 2 \cos^{2} \theta \ d \theta

    Use power reduction formulas:  \sin^{2} \theta = \frac{1- \cos 2 \theta}{2}

    and  \cos^{2} \theta = \frac{1+ \cos 2 \theta}{2}
    Quote Originally Posted by kittycat View Post
    hey turkeywilliam,

    I did. But I just can't get the right answer. I got (5pi)/64 +1/8 . But the correct answer is (5pi)/32. Don't know why?
     -\frac{1}{2} \int_{0}^{\pi/2}( (cos^2( \theta ) )^2 - 2 cos^2( \theta ))  d \theta

    = -\frac{1}{2} \int_{0}^{\pi/2}\left ( \left ( \frac{1 + cos(2 \theta )}{2} \right )^2  - 2 \left ( \frac{ 1 + cos(2 \theta )}{2} \right ) \right ) d \theta

    = -\frac{1}{8} \int_{0}^{\pi/2} (1 + cos(2 \theta ))^2 d \theta + \frac{1}{2} \int_0^{\pi/2} (1 + cos(2 \theta ) ) d \theta

    Expand the integrand in the first integral:
    = -\frac{1}{8} \int_{0}^{\pi/2}(1 + 2cos(2 \theta ) + cos^2(2 \theta) ) d \theta + \frac{1}{2} \int_0^{\pi/2} (1 + cos(2 \theta ) ) d \theta

    = \frac{3}{8} \int_{0}^{\pi/2} d \theta + \frac{1}{4} \int_0^{\pi/2} cos(2 \theta ) d \theta - \frac{1}{8} \int_0^{\pi/2} cos^2(2 \theta) d \theta

    Now use power reduction in the last integral again:
    = \frac{3}{8} \int_{0}^{\pi/2} d \theta + \frac{1}{4} \int_0^{\pi/2} cos(2 \theta ) d \theta - \frac{1}{8} \int_0^{\pi/2} \frac{1 + cos(4 \theta)}{2} d \theta

    = \frac{5}{16} \int_{0}^{\pi/2} d \theta + \frac{1}{4} \int_0^{\pi/2} cos(2 \theta ) d \theta - \frac{1}{16} \int_0^{\pi/2} cos(4 \theta) d \theta

    Now try it.

    -Dan
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