## Calculus, 10th Edition (Anton)

$$x\left( {\tan x - x} \right) + \ln \left| {\cos x} \right| + \frac{{{x^2}}}{2} + C$$
\eqalign{ & \int {x{{\tan }^2}x} dx \cr & {\text{identity 1 + ta}}{{\text{n}}^2}x = {\sec ^2}x \cr & = \int {x\left( {{{\sec }^2}x - 1} \right)} dx \cr & \cr & {\text{substitute }}u = x,{\text{ }}du = dx \cr & dv = \left( {{{\sec }^2}x - 1} \right)dx,{\text{ }} \cr & v = \int {\left( {{{\sec }^2}x - 1} \right)dx} \cr & v = \tan x - x \cr & {\text{ integration by parts}} \cr & \int {udv} = uv - \int {vdu} \cr & {\text{, we have}} \cr & = x\left( {\tan x - x} \right) - \int {\left( {\tan x - x} \right)dx} \cr & = x\left( {\tan x - x} \right) - \int {\tan xdx} + \int x dx \cr & {\text{trigonometric identity }}\tan x = \frac{{\sin x}}{{\cos x}} \cr & = x\left( {\tan x - x} \right) - \int {\frac{{\sin x}}{{\cos x}}dx} + \int x dx \cr & = x\left( {\tan x - x} \right) + \int {\frac{{ - \sin x}}{{\cos x}}dx} + \int x dx \cr & {\text{find antiderivative}} \cr & = x\left( {\tan x - x} \right) + \ln \left| {\cos x} \right| + \frac{{{x^2}}}{2} + C \cr}