That is the integral of du/(16-u^2)^(1/2) from 2 to 4?

Question

The following are posible answers: Pie/12 ; Pie/ 6 ; Pie/4 ; Pie/3; and 2Pie/3.......please show your work because I wan't to know how to do it....thank you

Answer 1

OK, whenever you have this sort of expression in the denominator you should be looking to do a trig substitution to take advantage of cos^2 x + sin^2 x = 1. So set u = 4 sin v, for instance. Then du = 4 cos v dv and (16-u^2)^(1/2) = (16(1-sin^2 v))^(1/2) = 4 cos v. So we get
int(2 to 4) du/(16-u^2)^(1/2)
= int (π/6 to π/2) (4 cos v dv / 4 cos v)
= int (π/6 to π/2) (dv)
= π/2 - π/6
= π/3.