Answer
$$\frac{{{{\tan }^2}t}}{2} + \ln \left| {\cos t} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^3}t} dt \cr
& {\text{Write the integrand as }}{\tan ^{2 + 1}}t \cr
& = \int {{{\tan }^{2 + 1}}t} dt \cr
& {\text{use the property }}{a^{m + }}^n = {a^m}{a^n} \cr
& = \int {{{\tan }^2}t\tan t} dt \cr
& {\text{use the trigonometric identity ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr
& = \int {\left( {{{\sec }^2}t - 1} \right)\tan t} dt \cr
& {\text{distribute}} \cr
& = \int {{{\sec }^2}t\tan t} dt - \int {\tan t} dt \cr
& {\text{integrate by using basic rules}} \cr
& = \frac{{{{\tan }^2}t}}{2} - \left( { - \ln \left| {\cos t} \right|} \right) + C \cr
& = \frac{{{{\tan }^2}t}}{2} + \ln \left| {\cos t} \right| + C \cr} $$