$$\int\sec^2(4x)\tan^2(4x)\,\mathrm{d}x$$
This is the original formula.
I used a U-substitution $u=4x$ so that means $ \frac{\mathrm{d}u}4=\mathrm{d}x $
So assuming I'm right then...
$$\frac14\int\sec^2(u)\tan^2(u)\,\mathrm{d}u$$
So I thought this would mean that after you take the integral you would have
$$-\frac14\frac{\tan^3(4x)\ln^3|4x|}{36}+C$$
However my webwork is telling me my answer is dead wrong. I believe it is because I messed up the product rule... but then I checked on a website but it wouldn't explain its answer with out money. So could some one work out the problem so I can see the proper integral? I am having trouble understanding how I could reverse the product rule. Do I have to use another substitution? Could I do this without substituting the trig function?