1. ## antiderivative w/ tan

Having trouble with this

find the antiderivative of
$\displaystyle \frac{1}{81z^2+16}$

2. ## Re: antiderivative w/ tan

We have something of the form $\displaystyle \frac{1}{a^2+x^2}$ which is a standard form that you can directly integrate.

If you don't want to do that, then we're going to have to do an icky substitution.

Let $\displaystyle z=\frac{4}{9}tan(x)$

Then $\displaystyle dz=\frac{4}{9}sec^2{x}~dx$

$\displaystyle I=\int\frac{1}{81z^2+16}dz$

$\displaystyle =\int\frac{\frac{4}{9}sec^2{x}}{81(\frac{16}{81}ta n^2(x))+16}dx$

I think it should work out. Good luck!