## Calculus: Early Transcendentals 8th Edition

$x tan (x) + ln|cos (x)| - x^2/2 + C$
Using integration by parts: u = x so u' = 1 v' = $tan^2 (x) = sec^2 (x) -1$ (Trigonometry identity) v = $\int sec^2 (x) -1$ dx = $tan(x) - x$ Now let's plug those parts into the equation: $\int x tan^2 (x) dx = uv - \int (u'v)$ =$x(tan(x) - x) - \int (tan(x) - x)dx$ = $x tan (x) - x^2 + ln|cos (x)| + (x^2)/2 + C$ = $x tan (x) + ln|cos (x)| - x^2/2 + C$