# Thread: Integrating (Inverse Trigonometric Functions)

1. ## Integrating (Inverse Trigonometric Functions)

Maybe I'm missing something obvious here, but I can't seem to integrate this. I'm just about positive I need to use arctan, but I can't get $\displaystyle a^2+u^2$ in the denominator.

$\displaystyle \int \frac{1}{3 + (x-2)^2}dx$

Any help is appreciated!

2. Let $\displaystyle u=x-2, \;\ du=dx$

$\displaystyle \int\frac{1}{3+u^{2}}du$

$\displaystyle \frac{1}{3}\int\frac{1}{1+(\frac{u}{\sqrt{3}})^{2} }du$

Now, use arctan.

3. Do you mind looking through to see if I did this correctly?

Using the formula for arctan:

$\displaystyle \int \frac{du}{a^2+u^2}=\frac{1}{a}arctan\frac{u}{a}+C$

And letting $\displaystyle a=1$, $\displaystyle u=\frac{x-2}{\sqrt{3}}$, and $\displaystyle du=\frac{\sqrt{3}}{3}$:

$\displaystyle \frac{1}{3}arctan\frac{x-2}{\sqrt{3}}+C$

Do you see anything I missed?

4. Looks good, except I think it should be $\displaystyle \frac{1}{\sqrt{3}}tan^{-1}(\frac{x-2}{\sqrt{3}})+C$