## Thomas' Calculus 13th Edition

$$3\pi - 8$$
\eqalign{ & \int_{ - \pi /4}^{\pi /4} {6{{\tan }^4}x} dx \cr & {\text{We write }}{\tan ^2}x{\text{ as ta}}{{\text{n}}^2}x{\tan ^2}x \cr & = 6\int {{\text{ta}}{{\text{n}}^2}x{{\tan }^2}x} dx \cr & {\text{use the fundamental identiy ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta - 1 \cr & = 6\int {\left( {{{\sec }^2}x - 1} \right){{\tan }^2}x} dx \cr & = 6\int {{{\sec }^2}x{{\tan }^2}x} dx - 6\int {{{\tan }^2}x} dx \cr & = 6\int {{{\sec }^2}x{{\tan }^2}x} dx - 6\int {\left( {{{\sec }^2}x - 1} \right)} dx \cr & = 6\int {{{\sec }^2}x{{\tan }^2}x} dx - 6\int {{{\sec }^2}x} dx + 6\int {dx} \cr & {\text{integrating}} \cr & = 2{\tan ^3}x - 6\tan x + 6x + C \cr & \cr & {\text{Then}} \cr & \int_{ - \pi /4}^{\pi /4} {6{{\tan }^4}x} dx = \left( {2{{\tan }^3}x - 6\tan x + 6x + C} \right)_{ - \pi /4}^{\pi /4} \cr & = \left( {2{{\tan }^3}\left( {\frac{\pi }{4}} \right) - 6\tan \left( {\frac{\pi }{4}} \right) + 6\left( {\frac{\pi }{4}} \right)} \right) - \left( {2{{\tan }^3}\left( { - \frac{\pi }{4}} \right) - 6\tan \left( { - \frac{\pi }{4}} \right) + 6\left( { - \frac{\pi }{4}} \right)} \right) \cr & {\text{simplifying}} \cr & = \left( {2 - 6 + \frac{3}{2}\pi } \right) - \left( { - 2 + 6 - \frac{3}{2}\pi } \right) \cr & = - 4 + \frac{3}{2}\pi - 4 + \frac{3}{2}\pi \cr & = 3\pi - 8 \cr}