integral 0 to sqrt(3) dx/sqrt(4-x^2)
I don't understand why the answer is pi/6. I get pi/3. Thanks for the help!
Work:
u=x/2 du=1/2dx 2du=dx
2/2sqrt(1-u^2)
sin^-1(x/2)
sin^-1(sqrt(3)/2)= pi/3
Integral of dx/sqrt(4-x^2
Let x = 2u and that becomes
2 du/sqrt(4 - 4u^2)
Integral of du/(sqrt(1-u^2)
= sin^1 u = sin^-1(x/2)
OK so far.. sin-1(sqrt3/2) - sin^-1 0
= pi/3
I agree with your answer