# How can I integrate $\int {dx \over \sqrt{3^2+x^2}}$ using Trigonometric Substitution?

$$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}}$$ $$x =3\tan\theta$$ $$dx = 3\sec^2\theta$$

$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$ $$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}} d\theta$$ $$\int {3\sec^2\theta \over 3\sec\theta} d\theta = \int \sec\theta$$ $$\ln|\sec\theta + \tan\theta| + C$$ $$\ln\left({\sqrt{9+x^2} \over 3} + {x \over 3}\right)$$

My book says the answer should just be:

$$\ln\left({\sqrt{9+x^2}} + {x}\right)$$

I'm wondering where I went wrong with this?

• To obtain $\sin x$, type \sin x in math mode. Similarly, type \cos x, \tan x, \sec x, \csc x, \cot x, \ln x, \log x in math mode to obtain $\cos x$, $\tan x$, $\sec x$, $\csc x$, $\cot x$, $\ln x$, and $\log x$, respectively. May 1, 2015 at 10:36

Hint: $$\log (t\times C) = \log t + \log C$$