Integration by parts 3

Submitted by roamingfree on Fri, 02/05/2021 - 11:08
skills
You need to be able to apply the by parts formula
Question

Evaluate the integral

$$ \int x\tan^2(x) dx $$
answer --- press button to toggle display
$ -x\tan(x) + \ln(\cos(x)) - \frac{1}{2}x^2 +c$
solution --- press button to toggle display
We need to apply the by parts formula. Let $u =x$ then $u' = 1$. Let $v' = \tan^2(x)$ then $v = \tan(x) - x$. $$ \begin{array}{rcl} \int x\tan^2(x) dx &=& x(\tan(x) - x) - \int 1(\tan(x) - x)dx \\ &=& -x\tan(x) - x^2 +\ln(\cos(x)) + \frac{1}{2}x^2 + c \\ &=& -x\tan(x) + \ln(\cos(x)) - \frac{1}{2}x^2 +c \end{array} $$