Question Number 31951 by NECx last updated on 17/Mar/18
$${Evaluate}\:\int\mathrm{sin}\:\sqrt{{x}}{dx} \\ $$
Answered by mrW2 last updated on 17/Mar/18
![u=(√x) du=(dx/(2(√x)))=(dx/(2u)) dx=2udu ∫sin (√x)dx =∫2u sin u du =−2∫ud cos u =−2[u cos u−∫cos u du] =−2[u cos u−sin u]+C =2[sin u−u cos u]+C =2[sin (√x)−(√x) cos (√x)]+C](https://www.tinkutara.com/question/Q31956.png)
$${u}=\sqrt{{x}} \\ $$$${du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}}}=\frac{{dx}}{\mathrm{2}{u}} \\ $$$${dx}=\mathrm{2}{udu} \\ $$$$\int\mathrm{sin}\:\sqrt{{x}}{dx} \\ $$$$=\int\mathrm{2}{u}\:\mathrm{sin}\:{u}\:{du} \\ $$$$=−\mathrm{2}\int{ud}\:\mathrm{cos}\:{u} \\ $$$$=−\mathrm{2}\left[{u}\:\mathrm{cos}\:{u}−\int\mathrm{cos}\:{u}\:{du}\right] \\ $$$$=−\mathrm{2}\left[{u}\:\mathrm{cos}\:{u}−\mathrm{sin}\:{u}\right]+{C} \\ $$$$=\mathrm{2}\left[\mathrm{sin}\:{u}−{u}\:\mathrm{cos}\:{u}\right]+{C} \\ $$$$=\mathrm{2}\left[\mathrm{sin}\:\sqrt{{x}}−\sqrt{{x}}\:\mathrm{cos}\:\sqrt{{x}}\right]+{C} \\ $$
Commented by mondodotto@gmail.com last updated on 19/Mar/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$
Commented by NECx last updated on 20/Mar/18
$${thanks}\:{so}\:{much}! \\ $$