I have the integration of radical(25 - x^2)/(x)
i set x = 5sinx
dx = 5cosx
i ended up with integral of cos^2(x)/sin(x)
..not so sure where to go from here. Thanks for the help.
$\displaystyle \int \frac{\sqrt{25 - x^2}}{x} dx$
Let $\displaystyle x = 5 sin(u)$ --> $\displaystyle dx = 5 cos(u) du$
Thus
$\displaystyle \int \frac{\sqrt{25 - x^2}}{x} dx = \int \frac{\sqrt{25 - 25sin^2(u)}}{5sin(u)} 5 cos(u) du$
$\displaystyle = 5 \sqrt{5} \int \frac{cos^2(u)}{sin(u)}du$
$\displaystyle = 5 \sqrt{5} \int \frac{cos^2(u) - 1 + 1}{sin(u)}du$
$\displaystyle = 5 \sqrt{5} \int \frac{cos^2(u) - 1}{sin(\theta)} du + 5 \sqrt{5} \int \frac{1}{sin(u)}du$
$\displaystyle = -5 \sqrt{5} \int sin(\theta) du + 5 \sqrt{5} \int \frac{1}{sin(u)}du$
Is that any better?
-Dan