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Math Help - integration

  1. #1
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    integration

    need a little help with this problem:
    \int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta
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  2. #2
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    Quote Originally Posted by viet View Post
    need a little help with this problem:
    \int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta
    The evaluation at the limits is simple once you find the anti-derivative. Thus, I will find the anti-derivative and let you do the evaluation.

    We have,
    \int \frac{\sin x}{1+\cos^2 x} dx
    Let u=\cos x then u'=-\sin x
    Substitution rule,
    - \int \frac{1}{1+u^2} du
    This, is the arctangent integral,
    -\tan^{-1}u+C=-\tan^{-1} (\cos x)+C=
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  3. #3
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    Quote Originally Posted by viet View Post
    need a little help with this problem:
    \int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta
    This will help:

    int(1/(1 + x^2)) = arctan(x)

    Let u = cos(x)

    Then,

    du = -sin(x)

    -du = sin(x)

    Factor out the -1

    -int(1/(1 + u)^2)du

    -arctan(u)du

    Thus, substitute in for u:

    -arctan(cos(x)) + C, but you're given the limits. And thus,

    -arctan(cos(Pi/2)) - [-arctan(cos(0))]

    0 - (-Pi/4)

    = Pi/4
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