Thread: Leibniz' pi integral

1. Leibniz' pi integral

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

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

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

Can somebody please show me how this integral is solved?

2. Originally Posted by espen180
Hi. I'm trying to do the integral Leibniz solved when he calculated pi. It goes as follows:

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

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

Can somebody please show me how this integral is solved?

Here's my stab at it:

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

Complete the square inside the radical:

$\displaystyle 2x-x^2=-(x^2-2x+1)+1=-(x-1)^2+1$

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

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

The limits of integration change as well!

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

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

Try to take it from here.

I hope this helps!

--Chris