integrate tan^2(3x)sec^4(3x)dx
integrate (1+sinx)/(cos^2 x) dx
$\displaystyle \int tan^{2}(3x)sec^{4}(3x)dx$
Let $\displaystyle u=3x, \;\ \frac{du}{3}=dx$
$\displaystyle \frac{1}{3}\int tan^{2}(u)sec^{4}(u)du$
Rewrite by using the identity $\displaystyle sec^{2}(u)=tan^{2}(u)+1$
$\displaystyle \frac{1}{3}\int sec^{2}(u)tan^{2}(u)du+\frac{1}{3}\int tan^{4}(u)sec^{2}(u)du$
Now, a simple sub and it works out well to finish.
Let $\displaystyle w=tan(u), \;\ dw=sec^{2}(u)du$
Then, resub.