sin-1-x-2-dx-

Question Number 93045 by john santu last updated on 10/May/20

$$\int\:\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{{x}}{\mathrm{2}}}\right)\:{dx}\: \\ $$

Commented by mathmax by abdo last updated on 10/May/20

I =∫ arcsin((√(x/2)))dx changement (√(x/2))=t give (x/2)=t^2 ⇒x=2t^2 I =∫ arcsin(t)(4t)dt =4 ∫ t arcsin(t)dt and by parts I =4{ (t^2 /2)arcsint −∫ (t^2 /2)(dt/( (√(1−t^2 ))))} =2t^2 arcsint −2∫ (t^2 /( (√(1−t^2 ))))dt ∫ (t^2 /( (√(1−t^2 ))))dt =_(t=sinα) ∫ ((sin^2 α)/(cosα)) cosα dα =∫ sin^2 α dα =(1/2)∫ (1−cos(2α))dα =(α/2) −(1/4)sin(2α)+c =(α/2)−(1/2)sinα cosα +c =(1/2)arcsint −(1/2)t (√(1−t^2 ))+c =(1/2)arcsin((√(x/2)))−(1/2)(√(x/2))(√(1−(x/2)))+c ⇒ I =xarcsin((√(x/2)))−arcsin((√(x/2)))+(√(x/2))(√(1−(x/2)))+c

$${I}\:=\int\:\:{arcsin}\left(\sqrt{\frac{{x}}{\mathrm{2}}}\right){dx}\:\:\:\:{changement}\:\sqrt{\frac{{x}}{\mathrm{2}}}={t}\:{give}\:\frac{{x}}{\mathrm{2}}={t}^{\mathrm{2}} \:\Rightarrow{x}=\mathrm{2}{t}^{\mathrm{2}} \\ $$$${I}\:=\int\:\:{arcsin}\left({t}\right)\left(\mathrm{4}{t}\right){dt}\:=\mathrm{4}\:\int\:{t}\:{arcsin}\left({t}\right){dt}\:\:\:{and}\:{by}\:{parts} \\ $$$${I}\:=\mathrm{4}\left\{\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}{arcsint}\:−\int\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\right\}\:=\mathrm{2}{t}^{\mathrm{2}} \:{arcsint}\:−\mathrm{2}\int\:\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$\int\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:=_{{t}={sin}\alpha} \:\:\int\:\:\frac{{sin}^{\mathrm{2}} \alpha}{{cos}\alpha}\:{cos}\alpha\:{d}\alpha\:=\int\:{sin}^{\mathrm{2}} \alpha\:{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\mathrm{1}−{cos}\left(\mathrm{2}\alpha\right)\right){d}\alpha\:=\frac{\alpha}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}\alpha\right)+{c} \\ $$$$=\frac{\alpha}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\alpha\:{cos}\alpha\:+{c}\:=\frac{\mathrm{1}}{\mathrm{2}}{arcsint}\:−\frac{\mathrm{1}}{\mathrm{2}}{t}\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{arcsin}\left(\sqrt{\frac{{x}}{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{{x}}{\mathrm{2}}}\sqrt{\mathrm{1}−\frac{{x}}{\mathrm{2}}}+{c}\:\Rightarrow \\ $$$${I}\:={xarcsin}\left(\sqrt{\frac{{x}}{\mathrm{2}}}\right)−{arcsin}\left(\sqrt{\frac{{x}}{\mathrm{2}}}\right)+\sqrt{\frac{{x}}{\mathrm{2}}}\sqrt{\mathrm{1}−\frac{{x}}{\mathrm{2}}}+{c} \\ $$

Commented by john santu last updated on 11/May/20

yes. i got = (x−1) sin^(−1) ((√(x/2)))+(1/2)(√(2x−x^2 ))+c

$$\mathrm{yes}.\:\mathrm{i}\:\mathrm{got}\: \\ $$$$=\:\left(\mathrm{x}−\mathrm{1}\right)\:\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{x}}{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2x}−\mathrm{x}^{\mathrm{2}} }+\mathrm{c}\: \\ $$

Answered by TANMAY PANACEA … last updated on 10/May/20

x=2sin^2 t dx=2×sin2t×dt ∫t×2sin2t dt (I/2)=∫t.sin2t dt =t∫sin2t dt−∫[(dt/dt)∫sin2t dt]dt =t×((cos2t)/(−2))−∫1×((cos2t)/(−2))dt =((−tcos2t)/2)+(1/2)×((sin2t)/2)+C =((−t)/2)×(1−2sin^2 t)+(1/4)×2sint×cost+c =((−sin^(−1) ((√(x/2)) ))/2)(1−x)+(1/2)×(√(x/2)) ×(1−(x/2))^(1/2) +c so I=−sin^(−1) ((√(x/2)) )(1−x)+(√(x/2)) (1−(x/2))^(1/2) +C_1

$${x}=\mathrm{2}{sin}^{\mathrm{2}} {t}\: \\ $$$${dx}=\mathrm{2}×{sin}\mathrm{2}{t}×{dt} \\ $$$$\int{t}×\mathrm{2}{sin}\mathrm{2}{t}\:{dt} \\ $$$$\frac{{I}}{\mathrm{2}}=\int{t}.{sin}\mathrm{2}{t}\:{dt} \\ $$$$={t}\int{sin}\mathrm{2}{t}\:{dt}−\int\left[\frac{{dt}}{{dt}}\int{sin}\mathrm{2}{t}\:{dt}\right]{dt} \\ $$$$={t}×\frac{{cos}\mathrm{2}{t}}{−\mathrm{2}}−\int\mathrm{1}×\frac{{cos}\mathrm{2}{t}}{−\mathrm{2}}{dt} \\ $$$$=\frac{−{tcos}\mathrm{2}{t}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{sin}\mathrm{2}{t}}{\mathrm{2}}+{C} \\ $$$$=\frac{−{t}}{\mathrm{2}}×\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {t}\right)+\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{2}{sint}×{cost}+{c} \\ $$$$=\frac{−{sin}^{−\mathrm{1}} \left(\sqrt{\frac{{x}}{\mathrm{2}}}\:\right)}{\mathrm{2}}\left(\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\frac{{x}}{\mathrm{2}}}\:×\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{c} \\ $$$${so}\:{I}=−{sin}^{−\mathrm{1}} \left(\sqrt{\frac{{x}}{\mathrm{2}}}\:\right)\left(\mathrm{1}−{x}\right)+\sqrt{\frac{{x}}{\mathrm{2}}}\:\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{C}_{\mathrm{1}} \\ $$

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