Find $\displaystyle \displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}$.
Is that what you are looking for?
$\displaystyle \displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}\times \sqrt{\sqrt{15}+2x}}{\sqrt{(2-x)(2+x)}(4-x^2)^2}\ dx$
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.
$\displaystyle \displaystyle \int_{-1}^{1}\frac{\sqrt{(\sqrt{15}-2x)(\sqrt{15}+2x)}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$
$\displaystyle \displaystyle \int_{-1}^{1}\frac{\sqrt{15 - 4x^2}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$
Can you continue now?
EDIT: Since the 'x' is an actual x, this becomes:
$\displaystyle \displaystyle \int_{-1}^{1}\frac{x\sqrt{15 - 4x^2}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$
$\displaystyle \displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{\sqrt{4-x^{2}}(4-x^2)^{2}}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.$
Do you notice anything about the integrand?
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
$\displaystyle \displaystyle \int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx=\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.$
Now, what happens when you flip the right integral's limits back to what they are on the LHS?
You are right. I forgot to account for the minus from the differential.
The integral on the right is
$\displaystyle \int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$
Flipping the limits to what they were
$\displaystyle \int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$
I get the integral in the left.
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 $\displaystyle \int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx = -\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$
So $\displaystyle \int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx = -\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$