Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
Don't tell me it is $2014$! –  Ian Mateus Jan 1 at 4:51
    
I don't think so. –  mathkiss Jan 1 at 5:01
    
Hint: you might start by completing the square. –  Robert Israel Jan 1 at 5:08
    
Now it is 2014! –  user21820 Jan 1 at 5:23

2 Answers 2

up vote 7 down vote accepted

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!

share|improve this answer

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?

share|improve this answer

Your Answer

 
discard

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.