Vtyjkyuk

$$intsec^2(4x)tan^2(4x),mathrmdx$$

This is the original formula.

I used a U-substitution $u=4x$ so that means $ fracmathrmdu4=mathrmdx $

So assuming I'm right then...

$$frac14intsec^2(u)tan^2(u),mathrmdu$$

So I thought this would mean that after you take the integral you would have

$$-frac14frac36+C$$

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?

$$frac14intsec^2(u)tan^2(u),mathrmdu$$
put $$tan(u)=t$$
thus $$sec^2(u),mathrmdu=mathrmdt$$
$$frac14int t^2,mathrmdt$$
$$fract^312+C$$
$$fractan^3(u)12+C$$
$$fractan^3(4x)12+C$$

Quick method:

$$
int tan^2(x)sec^2(x) mathrm dx = int tan^2(x) mathrm d(tan (x)) = cdots
$$

Hint:
$$frac14 int sec^2(u)tan^2(u) du=frac14 int (1+tan^2(u))tan^2(u) du$$
Now let $tan u=w$.