Alright, so my problem is problem #1 out of this homework section.

I have the integral of dx/(9+x^2)^(1/2).

Substituting,

x=3*tan*T

dx= 3sec^2t

I get that down to the integral of secT This is LN|SecU+TanU|+C.

Then I go to substitute back. I draw my right triangle. T starting from the origin, side opposite being along the X axis and positive.

So, tan T = x/3 . This means side opposite = X and side adjacent = 3. The hypotenuse will be (9+x^2)^(1/2).

Secant is hypotenuse over adjacent. So sec(t) = (9+x^2)^(1/2)/3 ...

Except the answer doesn't have 3s in the denominators. I should not have X/3 or (9+x^2)^(1/2)/3.

Where did I mess up?