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Question Number 170416 by Mastermind last updated on 23/May/22

∫(√(4−x^2 ))dx Mastermind

$$\int\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$$$ \\ $$$${Mastermind} \\ $$

Answered by thfchristopher last updated on 23/May/22

Let x=2sin t dx=2cos tdt sin t=(x/2), cos t=((√(4−x^2 ))/2) ∫(√(4−x^2 ))dx =∫(√(4−4sin^2 t ))(2cos tdt) =4∫cos^2 tdt =2∫(1+cos 2t)dt =2t+sin 2t+C =2t+2sin tcos t+C =2sin^(−1) ((x/2))+((x(√(4−x^2 )))/2)+C

$$\mathrm{Let}\:{x}=\mathrm{2sin}\:{t} \\ $$$${dx}=\mathrm{2cos}\:{tdt} \\ $$$$\mathrm{sin}\:{t}=\frac{{x}}{\mathrm{2}},\:\:\:\mathrm{cos}\:{t}=\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\int\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\int\sqrt{\mathrm{4}−\mathrm{4sin}^{\mathrm{2}} {t}\:}\left(\mathrm{2cos}\:{tdt}\right) \\ $$$$=\mathrm{4}\int\mathrm{cos}^{\mathrm{2}} {tdt} \\ $$$$=\mathrm{2}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}{t}+\mathrm{sin}\:\mathrm{2}{t}+{C} \\ $$$$=\mathrm{2}{t}+\mathrm{2sin}\:{t}\mathrm{cos}\:{t}+{C} \\ $$$$=\mathrm{2sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)+\frac{{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mathrm{2}}+{C} \\ $$

Commented by RANDHIR70 last updated on 23/May/22

Commented by Mastermind last updated on 24/May/22

Thanks

$$ \\ $$$${Thanks} \\ $$

Answered by RANDHIR70 last updated on 23/May/22

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