Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \int(225-u)^{1/2}\cdot u^{1/2}\, \mathrm du $$

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.

share|cite|improve this question
I have good news and bad news. The good news is that it's integrable. The bad news is that the integral looks mighty messy. If this is from a class, are you sure you copied it correctly? – mixedmath Aug 9 '11 at 16:28
Try $u=225\;\sin^2(t)$ and remember your double angle formulas. – robjohn Aug 9 '11 at 16:39
It would make your life easier if you replace 225 with $a^2$ and substitute $a=15$ in the final result. You could use Wolfram Alpha with input "int (a^2 - u)^(1/2) u^(1/2) du" and click "Show steps" button to see the chain of substitutions that lead to a table integral – Sasha Aug 9 '11 at 16:48
@Sasha : Wolfram Alpha gives the best automated integral explanations I've ever seen... it's a great way of stealing us reputation! (I say this as a compliment, just kidding) :P +1 – Patrick Da Silva Aug 9 '11 at 20:27
up vote 3 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
Why downvote? Is there smth incorrect in the answer? – Ilya Aug 9 '11 at 16:50
I noted the identity $$ \frac{(\sqrt{225-u} + \sqrt u)^2-225}2 = \sqrt{225-u}\sqrt{u} $$ But couldn't get anything out of it. Maybe you can? It seems nice to work with. – Patrick Da Silva Aug 9 '11 at 16:58
@Patrick Da Silva: are you sure? :) as for me, the first version is much more nice. – Ilya Aug 9 '11 at 17:25
I said I couldn't get anything out of it... how could I be sure? XD – Patrick Da Silva Aug 9 '11 at 17:59
SWP yields $8\int \sqrt{225u-u^{2}}du=4\sqrt{225u-u^{2}}u-450\sqrt{225u-u^{2}}+50625\arcsin \left( \frac{2}{225}u-1\right) $ – Américo Tavares Aug 9 '11 at 18:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.