# Evaluating $\int \sqrt{5 + 4x - x^2}dx$

$$\int \sqrt{5 + 4x - x^2}dx$$

I am pretty certain what I need to do to this problem is complete the square and turn it into a trig subsitution but I have no idea how to complete the square with a $-x^2$ or really with this problem at all, I just can't make it work.

I tried to see if I could make the problem be the same in any way by just pulling out a negative but that didn't seem to work.

I got the problem up to

$$\int \sqrt{ -1(x-2)^2 - 1}dx$$

But I do not think that does me any good. What I think I need to do is have a difference of squares with a square in it or something, I just have to get rid of the 4x term somehow.

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Don't forget the differentials. – Pedro Tamaroff Jun 9 '12 at 1:11
Should be $\int \sqrt{ 9-(x-2)^2}dx$ – GEdgar Jun 9 '12 at 1:16
Eugene: I think you mean $3\sin\theta=x-2$. – Cameron Buie Jun 9 '12 at 1:18

Firstly, it should be

$$\int \sqrt{5 + 4 + (-4) + 4x - x^2} dx = \int \sqrt{5 + 4 - (x^2 - 4x + 4}) dx = \int \sqrt{9 - (x - 2)^2}dx$$

Next a hint. Let $3\sin \theta = x-2$.

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When does this problems is it alright to leave the answer in terms of theta or whatever variable I use instead of x? I don't see why it matters if it is indefinite. – user138246 Jun 9 '12 at 1:30
Because it doesn't make much sense by the fundamental theorem of calculus that when you differentiate the left side you get a function in terms of $x$ and on the right you get a function in terms of $\theta$. – Eugene Jun 9 '12 at 1:33

Our integral can be written as, $$\begin{split} \int\sqrt{5+4x-x^{2}}dx&=\int\sqrt{9-4+4x-x^{2}}dx\\ &=\int\sqrt{3^{2}-(x-2)^{2}}dx.\ \end{split}$$ Now by trigonometric substitution, take $$x-2=3\sin \theta. (\because \text{If we have} \sqrt{a^{2}-x^{2}} \text{ then we have to substitute} x=a\sin\theta.)$$ Thus, $$\begin{split} \int\sqrt{3^{2}-(x-2)^{2}}dx&=\int\sqrt{3^{2}-3^{2}\sin^{2}\theta} 3\cos \theta d\theta\\ &=\int 3\sqrt{1-\sin^{2}\theta} 3\cos \theta d\theta\\ &=9\int\cos ^{2}\theta d\theta\\ &=9\int\frac{1+\cos 2\theta}{2}d\theta\\ &=\frac{9}{2}\int 1 d\theta+\frac{9}{2}\int \cos 2\theta d\theta\\ &=\frac{9}{2}\left[\theta+\frac{\sin 2\theta}{2}\right]+C=\frac{9}{2}\left[\theta +\frac{2\sin \theta\cos \theta}{2}\right]+C\\ &=\frac{9}{2}\left[\sin^{-1}\left(\frac{x-2}{3}\right)+\frac{1}{9}(x-2) \sqrt{9-(x-2)^{2}}\right]+C \end{split}$$ Thus, $$\int \sqrt{5+4x-x^{2}} dx=\frac{9}{2}\sin^{-1}\left(\frac{x-2}{3}\right)+\frac{(x-2) \sqrt{5+4x-x^{2}}}{2}+C$$

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I also have trouble completing the square if the coefficient of $x^2$ is negative. So I avoid doing it.

Let's look at our particular example $5+4x-x^2$. We have $$5+4x-x^2=-\left(x^2-4x-5\right).$$ Inside the parentheses, not only is the coefficient of $x^2$ positive, but $x^2$ is in front, where it likes to be. We are now in familiar territory, and can comfortably note that $$x^2-4x-5=(x-2)^2 -9.$$ Finally, take the negative of this. We get $9-(x-2)^2$. The rest has been well done by others: let $x-2=3\sin\theta$, or, more slowly, let $x-2=u$ and then let $u=3\sin\theta$.

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