# Math Help - Trig substitution integrals

1. ## [Solved]Trig substitution integrals

I solved the problem the 3's are part of the constant c thats why they arent in the answer.

Hello,
I have the integral

dy / sqrt( 9 + y^2)

I have worked it all the way down to

ln| sec theta + tan theta| + C

I need to now put it in terms of y.

from my previous work during the problem I have
y= 3 tan theta

so using right triangle trig
tan theta= y/3

hypotenuse is sqrt(9+y^2)

the answer I get is ln|(sqrt(9+y^2)/3) + (y/3)| + C
the book answer is ln|sqrt(9 + y^2) + y| + C

could someone please explain. I think I am having a misunderstanding with my substitutions.

2. $\ln\left|\frac{\sqrt{9+y^2}}{3} + \frac{y}{3}\right| + C$

$\ln\left|\frac{\sqrt{9+y^2} + y}{3}\right| + C$

$\ln\left|\sqrt{9+y^2} + y\right| - \ln{3} + C$

... $-\ln{3} + C$ is just another $C$

$\ln\left|\sqrt{9+y^2} + y\right| + C$