Question

$\int \left(1+x^2\right)\cos 2xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(1+x^2\right)·\cos 2x·dx \\ =\int \cos 2xdx+\int x^2·\cos 2xdx \\ =\frac{\sin 2x}{2}+I_{\mathit{1}}\mathrm{where}{I}_{\mathit{1}}=\int x^2\cos 2xdx...\left(1\right) \\ {I}_{\mathit{1}}=\int \underset{\mathrm{I}}{x^2}\underset{\mathrm{II}}{\cos 2x}dx \\ =x^2\int \cos 2xdx-\int \left\{\frac{d}{dx}\left(x^2\right)\int \cos 2xdx\right\}dx \\ =x^2·\frac{\sin 2x}{2}-\int \frac{2x\times \sin 2x}{2}dx \\ =\frac{x^2·\sin 2x}{2}-\int \underset{\mathrm{I}}{x}·\underset{\mathrm{II}}{\sin 2x}dx \\ =\frac{x^2·\sin 2x}{2}-\left[x\int \sin 2xdx-\int \left\{\frac{d}{dx}\left(x\right)\int \sin 2xdx\right\}dx\right] \\ =\frac{x^2·\sin 2x}{2}-\left[x\left(\frac{-\cos 2x}{2}\right)-\int 1·\left(\frac{-\cos 2x}{2}\right)dx\right] \\ =\frac{x^2·\sin 2x}{2}+\frac{x·\cos 2x}{2}-\frac{\sin 2x}{4}...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right) \\ \therefore I=\frac{\sin 2x}{2}+\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}+C \\ =\left(x^2+1\right)\frac{\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}+C \end{gathered}$$