# Math Help - Definite Integral

1. ## Definite Integral

Find $\displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}$.

2. Is that what you are looking for?

$\displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}\times \sqrt{\sqrt{15}+2x}}{\sqrt{(2-x)(2+x)}(4-x^2)^2}\ dx$

I only simplified the brackets and replace an 'x' by 'times', and simplified the denominator a little and added dx.

As you can see, the denominator and numerator can be further simplified.

$\displaystyle \int_{-1}^{1}\frac{\sqrt{(\sqrt{15}-2x)(\sqrt{15}+2x)}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$

$\displaystyle \int_{-1}^{1}\frac{\sqrt{15 - 4x^2}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$

Can you continue now?

EDIT: Since the 'x' is an actual x, this becomes:

$\displaystyle \int_{-1}^{1}\frac{x\sqrt{15 - 4x^2}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$

3. $\displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{\sqrt{4-x^{2}}(4-x^2)^{2}}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.$

Do you notice anything about the integrand?

4. @Unknown - It was an actual $x$. I'm sorry I should have made it clear.
Originally Posted by Ackbeet
$\displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{\sqrt{4-x^{2}}(4-x^2)^{2}}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.$

Do you notice anything about the integrand?
Looks reasonable now. I'll try to take it from there.

5. You should be able to write the answer down right now, with no further computation.

6. Ok, it just seemed strange that you'd have an x in the middle like that =P

7. Originally Posted by Ackbeet
You should be able to write the answer down right now, with no further computation.
You mean I don't need to find the antiderivative? That would be amazing, but I'm not sure how to do it. Is it to do with the domains of $\sqrt{15-4x^2}$ and $\sqrt{(4-x^2)^5}$?

8. You mean I don't need to find the antiderivative?
Not in this case. That's because the integrand plus your particular limits have a certain property that allows you to use a fantastic shortcut. What happens to the integrand if you replace x with -x?

9. Originally Posted by Ackbeet
Not in this case. That's because the integrand plus your particular limits have a certain property that allows you to use a fantastic shortcut. What happens to the integrand if you replace x with -x?
Limits flip and the sign changes - i.e. we get $\displaystyle -\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.
$

10. Actually, the limits flip, but the overall sign does NOT change. That's because you get one minus sign for the differential, and one minus sign for the lone x in the numerator. So you get

$\displaystyle \int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx=\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.$

Now, what happens when you flip the right integral's limits back to what they are on the LHS?

11. You are right. I forgot to account for the minus from the differential.
Originally Posted by Ackbeet
Now, what happens when you flip the right integral's limits back to what they are on the LHS?
The integral on the right is

$\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$

Flipping the limits to what they were

$\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$

I get the integral in the left.

12. No, no. What usually happens when you flip limits? (Scroll down the link to just below the Conventions section).

13. Originally Posted by Ackbeet
No, no. What usually happens when you flip limits? (Scroll down the link to just below the Conventions section).
Oh, I knew I was missing something.

So $\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx = -\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$

So $\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx = -\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$

14. So, if y = -y, what is y?

15. Originally Posted by Ackbeet
So, if y = -y, what is y?
$0$

That's it?

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