Answer
$$\frac{1}{2} + \ln \left( {\frac{{\sqrt 2 }}{2}} \right)$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^n}x} dx \cr
& {\text{*Rewrite the integrand}} \cr
& \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2}}x{{\tan }^2}x} dx \cr
& {\text{*Use the pythagorean identity }}{\tan ^2}x = {\sec ^2}x - 1 \cr
& \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)} dx \cr
& \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2}}x{{\sec }^2}x} dx - \int {{{\tan }^{n - 2}}x} \cr
& {\text{*Integrate}} \cr
& \int {{{\tan }^n}x} dx = \frac{{{{\tan }^{n - 2 + 1}}x}}{{n - 2 + 1}} - \int {{{\tan }^{n - 2}}x} \cr
& \int {{{\tan }^n}x} dx = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}x} \cr
& \cr
& {\text{Evaluate }}\int_0^{\pi /4} {{{\tan }^3}x} dx \cr
& \int_0^{\pi /4} {{{\tan }^3}x} dx = \left[ {\frac{{{{\tan }^{3 - 1}}x}}{{3 - 1}}} \right]_0^{\pi /4} - \int_0^{\pi /4} {\tan x} dx \cr
& \int_0^{\pi /4} {{{\tan }^3}x} dx = \left[ {\frac{{{{\tan }^2}x}}{2}} \right]_0^{\pi /4} + \left[ {\ln \left| {\cos x} \right|} \right]_0^{\pi /4} \cr
& \int_0^{\pi /4} {{{\tan }^3}x} dx = \left[ {\frac{{{{\tan }^2}\left( {\pi /4} \right)}}{2} - 0} \right] + \left[ {\ln \left| {\cos \frac{\pi }{4}} \right| - \ln \left| {\cos 0} \right|} \right] \cr
& {\text{Simplifying}} \cr
& \int_0^{\pi /4} {{{\tan }^3}x} dx = \frac{1}{2} + \ln \left( {\frac{{\sqrt 2 }}{2}} \right) \cr} $$
