Answer
$$\int 4\tan^3 2xdx=\tan^22x-2\ln|\sec2x|+C$$
Work Step by Step
$$A=\int 4\tan^3 2xdx$$
Use Reduction Formula 93, which states that
$$\int\tan^naxdx=\frac{\tan^{n-1}ax}{a(n-1)}-\int\tan^{n-2}axdx$$
for $n=3$ and $a=2$
$$A=4\Big[\frac{\tan^22x}{4}-\int\tan2xdx\Big]$$
Next, use Formula 89, which states that
$$\int\tan axdx=\frac{1}{a}\ln|\sec ax|+C$$
for $a=2$
$$A=4\Big[\frac{\tan^22x}{4}-\frac{1}{2}\ln|\sec2x|\Big]+C$$ $$A=\tan^22x-2\ln|\sec2x|+C$$
