$$\int\sec^2(4x)\tan^2(4x)\,\mathrm{d}x$$
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?
$\endgroup$ 43 Answers
$\begingroup$$$\frac14\int\sec^2(u)\tan^2(u)\,\mathrm{d}u$$ put $$\tan(u)=t$$ thus $$\sec^2(u)\,\mathrm{d}u=\mathrm{d}t$$ $$\frac14\int t^2\,\mathrm{d}t$$
$$\frac{t^3}{12}+C$$ $$\frac{\tan^3(u)}{12}+C$$ $$\frac{\tan^3(4x)}{12}+C$$
$\endgroup$ 4 $\begingroup$Quick method:
$$ \int \tan^2(x)\sec^2(x) \mathrm dx = \int \tan^2(x) \mathrm d(\tan (x)) = \cdots $$
$\endgroup$ $\begingroup$Hint:$$\frac{1}{4} \int \sec^2(u)\tan^2(u)\ du=\frac{1}{4} \int (1+\tan^2(u))\tan^2(u)\ du$$ Now let $\tan u=w$.
$\endgroup$ 2
