Question Number 35 by user2 last updated on 25/Jan/15
$$\mathrm{Evaluate}\:\:\int{x}^{\mathrm{2}} \mathrm{sin}\:{x}\:{dx} \\ $$
Answered by surabhi last updated on 04/Nov/14
![∫x^2 sin xdx= =x^2 ∫sin x dx−∫[(d/dx)(x^2 )∙∫sin x]dx =x^2 (−cos x)−∫2x(−cos x)dx =−x^2 cos x+2∫x cos x dx =−x^2 cos x+2[x(sin x)−∫{(d/dx)(x)∙∫cos x dx}dx] =−x^2 cos x+2[x sin x−∫sin x dx] =x^2 cos x+2[x sin x+cos x]+C](https://www.tinkutara.com/question/Q45.png)
$$\int{x}^{\mathrm{2}} \mathrm{sin}\:{xdx}= \\ $$$$={x}^{\mathrm{2}} \int\mathrm{sin}\:{x}\:{dx}−\int\left[\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \right)\centerdot\int\mathrm{sin}\:{x}\right]{dx} \\ $$$$={x}^{\mathrm{2}} \left(−\mathrm{cos}\:{x}\right)−\int\mathrm{2}{x}\left(−\mathrm{cos}\:{x}\right){dx} \\ $$$$=−{x}^{\mathrm{2}} \mathrm{cos}\:{x}+\mathrm{2}\int{x}\:\mathrm{cos}\:{x}\:{dx} \\ $$$$=−{x}^{\mathrm{2}} \mathrm{cos}\:{x}+\mathrm{2}\left[{x}\left(\mathrm{sin}\:{x}\right)−\int\left\{\frac{{d}}{{dx}}\left({x}\right)\centerdot\int\mathrm{cos}\:{x}\:{dx}\right\}{dx}\right] \\ $$$$=−{x}^{\mathrm{2}} \mathrm{cos}\:{x}+\mathrm{2}\left[{x}\:\mathrm{sin}\:{x}−\int\mathrm{sin}\:{x}\:{dx}\right] \\ $$$$={x}^{\mathrm{2}} \mathrm{cos}\:{x}+\mathrm{2}\left[{x}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right]+{C} \\ $$