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Math Help - can someone confirm this

  1. #1
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    can someone confirm this

    the integral from 0 to pi/2 of [(sinx)/(cos(x)+1)]

    is -ln lcosx+1l + c
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  2. #2
    Super Member Showcase_22's Avatar
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    \int_0^{\frac{\pi}{2}} \frac{ \sin x}{\cos x +1}~dx

    = \left[ - \ln| \cos x+1| \right]_0^{\frac{\pi}{2}}

    =- \ln 1+ \ln 2= \ln 2

    You forgot to put in your limits!
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    \int_0^{\frac{\pi}{2}} \frac{ \sin x}{\cos x +1}~dx

    = \left[ - \ln| \cos x+1| \right]_0^{\frac{\pi}{2}}

    =- \ln 1+ \ln 2= \ln 2

    You forgot to put in your limits!
    so its just ln2?
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  4. #4
    Super Member Showcase_22's Avatar
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    Yes! What you evaluated was the integral without limits (or the "indefinite integral").

    When you have limits you can just drop the +c and start substituting.
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  5. #5
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    Quote Originally Posted by Showcase_22 View Post
    Yes! What you evaluated was the integral without limits (or the "indefinite integral").

    When you have limits you can just drop the +c and start substituting.
    i dont understand how a -ln1+ln2=ln2. shouldnt it just be pos ln1?
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  6. #6
    Super Member Showcase_22's Avatar
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    Well \ln 1=0 so you can just get rid of it.

    Alternatively, use the laws of logs:

    \ln 2 - \ln 1=\ln \frac{2}{1}= \ln 2
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  7. #7
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    Quote Originally Posted by Showcase_22 View Post
    Well \ln 1=0 so you can just get rid of it.

    Alternatively, use the laws of logs:

    \ln 2 - \ln 1=\ln \frac{2}{1}= \ln 2
    thanks
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