# Thread: Help with this arctan integration

1. ## Help with this arctan integration

Hello guys. So, I end up facing this integral in a homework, and it looked fine to me. To be honest, I think I have the right answer, but the computer keeps saying it's wrong. I am not sure where's my mistake.

The answer I got was $\displaystyle 9/4 arctan (x/4) + C$. I have absolutely no idea what I did wrong

Thanks for the help guys!

2. ## Re: Help with this arctan integration

Let $\displaystyle x^2 = 16\tan^2\theta$. Then $\displaystyle x = 4\tan \theta$, so $\displaystyle dx = 4\sec^2\theta d\theta$.

$\displaystyle \int \dfrac{9}{\sqrt{x^2+16}}dx = \int \dfrac{36\sec^2\theta d\theta}{\sqrt{16\sec^2\theta}} = 9\int \sec \theta d\theta$

The integral of $\displaystyle \sec \theta$ is $\displaystyle \ln\left|\sec \theta + \tan \theta\right|+C$. Since $\displaystyle \tan \theta = \dfrac{x}{4} = \dfrac{\text{opp}}{\text{adj}}$, you know $\displaystyle \sec \theta = \dfrac{\text{hyp}}{\text{adj}} = \dfrac{\sqrt{x^2+16}}{4}$. So, the integral is:

$\displaystyle \int \dfrac{9}{\sqrt{x^2+16}}dx = 9\ln\left|\dfrac{\sqrt{x^2+16}}{4} + \dfrac{x}{4}\right|+C = 9\ln\left|\sqrt{x^2+16}+x\right|+C$

3. ## Re: Help with this arctan integration

Thak you so much! Much appreciated!