Question

$\int x^2\cos 2xdx$

Answer 1

$$\begin{gathered} \int x^2\cos 2xdx \\ \mathrm{Taking}{x}^2\mathrm{as}\mathrm{the}\mathrm{first}\mathrm{function}\mathrm{and}\cos 2x\mathrm{as}\mathrm{the}\mathrm{second}\mathrm{function}. \\ =x^2\int \cos 2xdx-\int \left(2x\int \cos 2xdx\right)dx \\ =\frac{x^2\sin 2x}{2}-\int \frac{2x\sin 2x}{2}dx \\ =\frac{x^2}{2}\sin 2x-\int x\sin 2xdx \\ =\frac{x^2}{2}\sin 2x-\left[x\int \sin 2x-\int \left(\int \sin 2xdx\right)dx\right] \\ =\frac{x^2}{2}\sin 2x-\left[\frac{-x\cos 2x}{2}+\int \frac{\cos 2x}{2}dx\right] \\ =\frac{x^2}{2}\sin 2x+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}+C \end{gathered}$$