Having trouble with this
find the antiderivative of
$\displaystyle \frac{1}{81z^2+16}$
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!