Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 31951 by NECx last updated on 17/Mar/18

Evaluate ∫sin (√x)dx

$${Evaluate}\:\int\mathrm{sin}\:\sqrt{{x}}{dx} \\ $$

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

$${u}=\sqrt{{x}} \\ $$$${du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}}}=\frac{{dx}}{\mathrm{2}{u}} \\ $$$${dx}=\mathrm{2}{udu} \\ $$$$\int\mathrm{sin}\:\sqrt{{x}}{dx} \\ $$$$=\int\mathrm{2}{u}\:\mathrm{sin}\:{u}\:{du} \\ $$$$=−\mathrm{2}\int{ud}\:\mathrm{cos}\:{u} \\ $$$$=−\mathrm{2}\left[{u}\:\mathrm{cos}\:{u}−\int\mathrm{cos}\:{u}\:{du}\right] \\ $$$$=−\mathrm{2}\left[{u}\:\mathrm{cos}\:{u}−\mathrm{sin}\:{u}\right]+{C} \\ $$$$=\mathrm{2}\left[\mathrm{sin}\:{u}−{u}\:\mathrm{cos}\:{u}\right]+{C} \\ $$$$=\mathrm{2}\left[\mathrm{sin}\:\sqrt{{x}}−\sqrt{{x}}\:\mathrm{cos}\:\sqrt{{x}}\right]+{C} \\ $$

Commented by mondodotto@gmail.com last updated on 19/Mar/18

thank you very much

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

Commented by NECx last updated on 20/Mar/18

thanks so much!

$${thanks}\:{so}\:{much}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com