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Math Help - Leibniz' pi integral

  1. #1
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    Leibniz' pi integral

    Hi. I'm trying to do the integral Leibniz solved when he calculated pi. It goes as follows:

    I=\int_0^1 \sqrt{2x-x^2}\rm{d}x=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2n-1}=\frac{\pi}{4}

    I know the solution to the integral, but I can't solve it.

    Can somebody please show me how this integral is solved?

    Thanks in advance.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by espen180 View Post
    Hi. I'm trying to do the integral Leibniz solved when he calculated pi. It goes as follows:

    I=\int_0^1 \sqrt{2x-x^2}\rm{d}x=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2n-1}=\frac{\pi}{4}

    I know the solution to the integral, but I can't solve it.

    Can somebody please show me how this integral is solved?

    Thanks in advance.
    Here's my stab at it:

    \int_0^1\sqrt{2x-x^2}\,dx

    Complete the square inside the radical:

    2x-x^2=-(x^2-2x+1)+1=-(x-1)^2+1

    Thus, the integral becomes \int_0^1\sqrt{1-(x-1)^2}\,dx

    Now apply the trig substitution x-1=\sin\vartheta. We then see that \,dx=\cos\vartheta\,d\vartheta

    The limits of integration change as well!

    I leave it for you to verify that the integral becomes \int_{-\frac{\pi}{2}}^0\cos^2\vartheta\,d\vartheta

    Note that \cos^2\vartheta=\tfrac{1}{2}(1+\cos(2\vartheta))

    Try to take it from here.

    I hope this helps!

    --Chris
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