Blog: `int x tan^2 x dx` Evaluate the integral

Sunday, November 23, 2014

$\displaystyle\int{x}{{\tan}}^{{2}}{x}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int{x}{{\tan}}^{{2}}{x}{\left.{d}{x}\right.}$

Rewrite the integrand using the identity $\displaystyle{{\tan}}^{{2}}{x}={{\sec}}^{{2}}{x}-{1}$

$\displaystyle\int{x}{{\tan}}^{{2}}{x}{\left.{d}{x}\right.}=\int{x}{\left({{\sec}}^{{2}}{x}-{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int{x}{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}-\int{x}{\left.{d}{x}\right.}$

Now let's evaluate $\displaystyle\int{x}{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}$ using integration by parts,

$\displaystyle\int{x}{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}={x}\cdot\int{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}-\int{\left(\frac{{d}}{{\left.{d}{x}}}{\left({x}\right)}\int{{\sec}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle={x}{\tan{{\left({x}\right)}}}-\int{\left({1}\cdot{\tan{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle={x}{\tan{{\left({x}\right)}}}-\int{\left(\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}}\right)}{\left.{d}{x}\right.}$

Substitute cos(x)=t

-sin(x)dx=dt

$\displaystyle\int{\left(\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}}\right)}{\left.{d}{x}\right.}=\int-\frac{{\left.{d}{t}}}{{t}}$

$\displaystyle=-{\ln}{\left|{t}\right|}$

substitute back t=cos(x),

$\displaystyle=-{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}$

$\displaystyle\int{x}{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}={x}{\tan{{\left({x}\right)}}}-{\left(-{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}\right)}$

$\displaystyle={x}{\tan{{\left({x}\right)}}}+{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}$

$\displaystyle\int{x}{{\tan}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}={x}{\tan{{\left({x}\right)}}}+{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}-\int{x}{\left.{d}{x}\right.}$