Hi.

This is a definite integration question [0.1/10]

The questions is = {1 / root1-3x^2)}dx

I did first

for the integrand 1 / root(1-3x^2), substitute u = root3 x , du = root3 dx ..

= 1/root3 * 1/root(1-u^2) du

1/root(1-u^2) = sin^1(u)

Then, [sin^-1(u) / root3] 1~1/10

= [sin^-1(root3 x) / root3] 1~ 1/10

= sin^-1(root3 /10) / root3

May I ask when I am wrong?

Please help me out