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.
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Yes, so if you define $\displaystyle x=4\sin u,\,-\frac\pi2<u<\frac\pi2\implies dx=4\cos u\,du.$ And $\displaystyle 16-x^2=16-16\sin^2u=16\cos^2u.$ You can do the rest from there.
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