Alright, so my problem is problem #1 out of this homework section.
I have the integral of dx/(9+x^2)^(1/2).
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?