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Math Help - trig function

  1. #1
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    trig function

    Hello all

    here is my problem

     \int {\sqrt{1+\sin x}\,dx}

    i tired a half angle approach ( and didnt get very far). any hints?
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  2. #2
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    Quote Originally Posted by bobak View Post
    Hello all

    here is my problem

     \int {\sqrt{1+\sin x}\,dx}

    i tired a half angle approach ( and didnt get very far). any hints?
    Hint only?

    cos^2(X) = 1 -sin^2(X)
    so,
    cosX = sqrt[1 -sin^2(X)]
    cosX = sqrt[(1 +sinX)(1 -sinX)]

    Further,
    INT.[sqrt(1 +sinX)]dX
    = INT.[sqrt(1 +sinX)]*[cosX / cosX] dX
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  3. #3
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    -2\sqrt{1-\sin x} that right ?
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  4. #4
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    Yes.

    But you have to add the all-impotant "C".

    = -2sqrt(1 -sinX) +C
    Last edited by ticbol; October 22nd 2007 at 12:30 PM.
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  5. #5
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    Hello,bobak!

    Here is my problem: .  \int \sqrt{1+\sin x}\,dx
    Multiply by \frac{1-\sin x}{1-\sin x}

    . . \sqrt{\frac{1-\sin x}{1-\sin x}\cdot(1+\sin x)}\;=\;\sqrt{\frac{1-\sin^2x}{1-\sin x}} \;=\;\sqrt{\frac{\cos^2x}{1-\sin x}}\;=\;\frac{\cos x}{\sqrt{1-\sin x}}


    We have: . \int\frac{\cos x}{\sqrt{1-\sin x}}\,dx \;=\;\int(1-\sin x)^{-\frac{1}{2}}(\cos x\,dx)


    Now let: u \,= \,1-\sin x

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