Is 1-3x^2 all under the square root? Sorry, I can see it is.
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