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

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.

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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
4  
Try $u=225\;\sin^2(t)$ and remember your double angle formulas. –  robjohn Aug 9 '11 at 16:39
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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

2 Answers 2

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.

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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.

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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

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