Answer
$\dfrac{x^2}{2}\sin (2x)+\dfrac{x}{2} \cos (2x)-\dfrac{1}{4} \sin (2 x) +C$
Work Step by Step
Our aim is to solve the integral $ \int x^2 \cos (2x) \ dx$
Use integration by parts formula.
$ \int x^2 \cos (2x) =\dfrac{x^2}{2} \times \sin (2x) -\int x (\sin 2x) \ dx $
or, $=\dfrac{x^2}{2} \times \sin (2x) -[-\dfrac{x}{2} (\cos 2x)+\dfrac{1}{2} \int \cos (2 x) \ dx $
or, $=\dfrac{x^2}{2}\sin (2x)+\dfrac{x}{2} \cos (2x)-\dfrac{1}{4} \sin (2 x) +C$
