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Math Help - integration - please help!

  1. #1
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    Question integration - please help!

    2 times the integral ( from 0 to pi/2) of [ sin theta (1+cos theta)^2 -sin theta ] d(theta).

    Thank you very much.
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  2. #2
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    Quote Originally Posted by kittycat View Post
    2 times the integral ( from 0 to pi/2) of [ sin theta (1+cos theta)^2 -sin theta ] d(theta).

    Thank you very much.
    This is
    2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 - sin(\theta)] d\theta
    right?

    = 2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 ] d \theta - 2 \int_0^{\pi/2} sin(\theta) d\theta

    The second integral is trivial. For the first integral, let
    u = 1 + cos(\theta) \implies dx = -sin(\theta) d\theta

    So the first integral becomes:
    = 2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 ] d \theta = 2 \int_2^1 u^2 \cdot (-1) du

    = -2 \int_2^1 u^2 du = 2 \int_1^2 u^2 du
    which is another trivial integral.

    -Dan
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  3. #3
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    One approach may be to rewrite as:

    sin{\theta}(1+cos{\theta})^{2}-sin{\theta}=

    \frac{sin(3\theta)}{4}+sin(2\theta)+\frac{sin(\the  ta)}{4}

    Makes for an easy term by term integration.
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