1. Evaluate the Integral

The question says

Evaluate

(10x^2) dx/ sqroot 16-x^2

from 0 to 2 sqroot 3

I know this is a trig substitution for sin, but I'm having problems working out the trig identities.

2. Yes, so if you define $x=4\sin u,\,-\frac\pi2

And $16-x^2=16-16\sin^2u=16\cos^2u.$

You can do the rest from there.