Calculus
posted by Anonymous .
integral 0 to sqrt(3) dx/sqrt(4x^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(1u^2)
sin^1(x/2)
sin^1(sqrt(3)/2)= pi/3

Integral of dx/sqrt(4x^2
Let x = 2u and that becomes
2 du/sqrt(4  4u^2)
Integral of du/(sqrt(1u^2)
= sin^1 u = sin^1(x/2)
OK so far.. sin1(sqrt3/2)  sin^1 0
= pi/3
I agree with your answer