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Thread: How do I solve this [tex] \int_\frac{pi}{2}^0 \frac{cosx}{1 + sin^2x} dx[/tex]?

  1. #1
    Junior Member bijosn's Avatar
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    How do I solve this [tex] \int_\frac{pi}{2}^0 \frac{cosx}{1 + sin^2x} dx[/tex]?

    How do I solve this  \int_0^\frac{pi}{2} \frac{cosx}{1 + sin^2x} dx
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  2. #2
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    Make the substitution \displaystyle u = \sin{x} \implies du = \cos{x}\,dx.
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  3. #3
    Junior Member bijosn's Avatar
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    Quote Originally Posted by Prove It View Post
    Make the substitution \displaystyle u = \sin{x} \implies du = \cos{x}\,dx.

    \int \frac{du}{1 + u^2}
    =tan^{-1}u
    =tan^{-1}(sin x)\big|_0^{\frac{pi}{2}}

    am I doing this right?
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  4. #4
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    Yes, now substitute the terminals...
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  5. #5
    Junior Member bijosn's Avatar
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    \int \frac{du}{1 + u^2}
    =tan^{-1}u
    =tan^{-1}(sin x)\big|_0^{\frac{pi}{2}}
    =tan^{-1}(1) - tan^{-1}(0)
    like this?

    = pi/4
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  6. #6
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    Correct
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  7. #7
    Junior Member bijosn's Avatar
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    Quote Originally Posted by Prove It View Post
    Correct
    great success thank you again
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