# Integral: $I=\int\limits_{0}^{1}\dfrac{x^2dx}{\sqrt{3+2x-x^2}}$

Evaluate: $\displaystyle I=\int\limits_{0}^{1}\dfrac{x^2dx}{\sqrt{3+2x-x^2}}$

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Don't tell me it is $2014$! – Ian Mateus Jan 1 '14 at 4:51
I don't think so. – mathkiss Jan 1 '14 at 5:01
Hint: you might start by completing the square. – Robert Israel Jan 1 '14 at 5:08
Now it is 2014! – user21820 Jan 1 '14 at 5:23

The tricky part is the integration. I'll leave the limits for you to deal with. So we are looking at $$\int \frac{x^2}{\sqrt{3+2x-x^2}} \;dx$$ First, complete the square in the denominator to obtain....

$$\int \frac{x^2}{\sqrt{4-(x-1)^2}} \;dx$$

Then we make a substitution. What 'looks good' here?

Let $u=x-1$ and $du=dx$, then we have $$\int \frac{(u+1)^2}{\sqrt{4-u^2}}\;du$$

Then the 'easy' way is to expand the numerator and obtain

$$\int \frac{u^2}{\sqrt{4-u^2}}\;du +\int \frac{2u}{\sqrt{4-u^2}} \;du+\int \frac{1}{\sqrt{4-u^2}}\;du$$

Which are three simple integrations done by trig substitution, $u$-substitution, and special formulas, respectively. One could also have done a substitution at the previous step of $u=2\sin v$ then expand and integrate term by term. In any case, it's messy and time consuming but not difficult. Good luck!

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As $\displaystyle\frac{x^2}{3+2x-x^2}=\frac{x^2}{2^2-(x-1)^2}$ and we are dealing with definite integral

I would like to use Trigonometric substitution $\displaystyle x-1=2\sin u$ from the very start as we don't need to get back to $x$ again

When $\displaystyle x=0,\sin u=-\frac12\implies u=-\frac\pi6$ and when $\displaystyle x=1,\sin u=0\implies u=0$ based on the definition of the principal value of sine inverse

$\displaystyle\implies\cos u>0$

$$\int_0^1\frac{x^2}{\sqrt{3+2x-x^2}}dx=\int_0^{-\frac\pi6}\frac{(1+2\sin u)^2}{+2\cos u}2\cos udu=\int_0^{-\frac\pi6}(1+2\sin u)^2du$$

Now, $\displaystyle(1+2\sin u)^2=1+4\sin u+4\sin^2u=1+4\sin u+2(1-\cos2u)$ (using $\cos2A=1-2\sin^2A$)

$\displaystyle\implies(1+2\sin u)^2=3+4\sin u-2\cos2u$

Can you take it home from here?

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