Question

$\int x^2{\sin }^{2}xdx$

Answer 1

$$\begin{gathered} \int x^2{\sin }^{2}xdx \\ \mathrm{Taking}{x}^2\mathrm{as}\mathrm{the}\mathrm{first}\mathrm{function}\mathrm{and}{\sin }^{2}x\mathrm{as}\mathrm{the}\mathrm{second}\mathrm{function}. \\ =x^2\int \frac{1-\cos 2x}{2}-\int \left\{\frac{d}{dx}\left(x^2\right)\int \frac{1-\cos 2x}{2}dx\right\}dx \\ =\frac{x^2}{2}\left[x-\frac{\sin 2x}{2}\right]-\int 2x\left(\frac{x-{\displaystyle \frac{\sin 2x}{2}}}{2}\right)dx \\ =\frac{x^2}{2}\left(x-\frac{\sin 2x}{2}\right)-\int x^{2}dx+\int \frac{x\sin 2x}{2}dx\left[\mathrm{Here},\mathrm{taking}x\mathrm{as}\mathrm{the}\mathrm{first}\mathrm{function}\mathrm{and}\sin 2x\mathrm{as}\mathrm{the}\mathrm{second}\mathrm{function}.\right] \\ =\frac{x^3}{2}-\frac{x^2\sin 2x}{4}-\frac{x^3}{3}+\frac{1}{2}\left[x\int \sin 2x-\int \left\{\frac{d}{dx}\left(x\right)\int \sin 2xdx\right\}dx\right] \\ =\frac{x^3}{2}-\frac{x^2\sin 2x}{4}-\frac{x^3}{3}+\frac{1}{2}\left[\frac{-x\cos 2x}{2}+\int \frac{\cos 2xdx}{4}\right] \\ =\frac{x^3}{6}-\frac{x^2\sin 2x}{4}-\frac{x\cos 2x}{4}+\frac{\sin 2x}{8}+C \end{gathered}$$