I thought it would be fun to see if I could find the area of a circle with a radius of 5 by taking the integral of the positive form of a circle equation and multiplying it by two (since I believe the area of the bottom half is equal to the area of the upper half, or I should be able to take a definite integral from 0 to 5 and multiply it by 4). It turns out that integrating it was not as easy as I thought it would be.

$\displaystyle \int{\sqrt{25-x^2}}\text{ }dx$

I tried u-replacement, even though it looks like nothing will cancel.

$\displaystyle \int{\sqrt{u}}\text{ }dx$

$\displaystyle u = 25-x^2$

$\displaystyle \frac{du}{dx}=-2x$

$\displaystyle du = -2x*dx$

$\displaystyle dx = -\frac{1}{2x}*du$

$\displaystyle \int{\sqrt{u}}*\frac{1}{2x}\text{ }du$

$\displaystyle u = 25-x^2$

$\displaystyle x = \sqrt{25-u}$

$\displaystyle \int{\sqrt{u}}*\frac{1}{2\sqrt{25-u}}\text{ }du$

I don't think there's anything I can do with this.

It looks like the equation fits one of the inverse trigonomic integrals, but it's not a fraction.

Next, I try to integrate by parts. I can't really think of anything else.

$\displaystyle \int{\sqrt{25-x^2}}\text{ }dx = \int{\sqrt{(5+x)}*\sqrt{(5-x)}}$

By the way, if there's something I should have seen here that would indicate to me that integrating by parts here will not work, let me know.

$\displaystyle u = \sqrt{5-x}$

$\displaystyle v = \sqrt{5+x}$

$\displaystyle \frac{du}{dx} = \frac{1}{2}(5-x)^{-\frac{1}{2}} * -1$

$\displaystyle \int{\sqrt{5+x}} = \frac{2}{3}(5+x)^{\frac{3}{2}}$

$\displaystyle \int{\sqrt{(5+x)} * \sqrt{(5-x)}} =\sqrt{5-x} * \frac{2}{3}(5+x)^{\frac{3}{2}} - \int{\frac{2}{3}(5+x)^{\frac{3}{2}} * \frac{1}{2}(5-x)^{-\frac{1}{2}} * -1}$

I don't think it helped. I could maybe integrate by parts again, but it would probably not make it any easier to solve. What could I do to solve this?