what is the integral of the given radicals $$
\int(225-u)^{1/2}\cdot u^{1/2}\, \mathrm du
$$
I don't think it is allowed to combine both because of the minus sign, that's why I'm at a loss on how to integrate this. Please help.
 A: First complete the square:
$$
\int \sqrt{225 -u}\;\sqrt{u}\;du = \int \sqrt{225u -u^2}\;du = \int \sqrt{\left(\frac{225}{2}\right)^2 - \left(u^2 - 225u + \left(\frac{225}{2}\right)^2\right)} \; du
$$
$$
= \int\sqrt{\left(\frac{225}{2}\right)^2 - \left(u - \frac{225}{2}\right)^2} \; du = \frac{225}{2} \int\sqrt{1 - \left(\text{something}\right)^2} \; du
$$
Then let $\sin\theta = \text{something}$ and differentiate in order to figure out what goes in place of $du$.
A general idea is that where you have a quadratic polynomial with a first-degree term (in this case the polynomial is $225u - u^2$) you can reduce it to a quadratic polynomial with no first-degree term by completing the square.  That's what completing the square is for.
A: Sorry for the incomplete answer. This is a Chebyshev integral of the kind
$$
\int\limits x^m(ax^n+b)^p\,dx
$$
and there is a theorem that you can express it in elementary functions iff $p\in \mathbb Z$ or $\frac{m+1}{n}\in\mathbb Z$ or $p+\frac{m+1}{n}\in \mathbb Z$. In each case there are substitutions which allow you to express it through the elementary functions.
In your case $m=0.5,n=1$ and $p=0.5$, so $p+\frac{m+1}{n} = 2\in\mathbb Z$ and there is a solution. Unfortunately, I couldn't find substitutions - whenever I will find it, I will put it here.
