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Question Number 28887 by abdo imad last updated on 31/Jan/18

find ∫ arcsin((√(x/(x+2))))dx.

$${find}\:\int\:\:{arcsin}\left(\sqrt{\frac{{x}}{{x}+\mathrm{2}}}\right){dx}. \\ $$

Commented by abdo imad last updated on 02/Feb/18

let use the ch.arcsin((√( (x/(x+1)))) )=t ⇔(√(x/(x+1))) =sint ⇔(x/(x+1))=sin^2 t ⇔x=xsin^2 t +sin^2 t⇔(1−sin^2 t)x =sin^2 t⇔ x= ((sin^2 t)/(1−sin^2 t))=− ((1−sin^2 t −1)/(1−sin^2 t))=−1 −(1/(sin^2 t−1))⇒ dx=((2sint cost)/((sin^2 t −1)^2 ))dt so I= ∫ t ((2sint cost)/((1−sin^2 t)^2 ))dt = ∫ ((2t sint cost)/(cos^4 t))dt = ∫ ((2t sint)/(cos^3 t))dt by parts u=2t and v^′ = ((sint)/(cos^3 t)) I= −t cos^(−2) t −∫ 2(((−1)/2)cos^(−2) t)dt = −(t/(cos^2 t)) + ∫ (dt/(cos^2 t))= tant −(t/(cos^2 t)) +k but cos^2 t =1−sin^2 t=1−(x/(x+1))= (1/(x+1)) and tant= ((sint)/(cost))=((√(x/(x+1)))/(√(1/(x+1))))= x so I= (√x)−(1/(x+1))arcsin((√(x/(x+1))) ) +k .

$${let}\:{use}\:{the}\:{ch}.{arcsin}\left(\sqrt{\:\frac{{x}}{{x}+\mathrm{1}}}\:\right)={t}\:\Leftrightarrow\sqrt{\frac{{x}}{{x}+\mathrm{1}}}\:={sint} \\ $$$$\Leftrightarrow\frac{{x}}{{x}+\mathrm{1}}={sin}^{\mathrm{2}} {t}\:\Leftrightarrow{x}={xsin}^{\mathrm{2}} {t}\:+{sin}^{\mathrm{2}} {t}\Leftrightarrow\left(\mathrm{1}−{sin}^{\mathrm{2}} {t}\right){x}\:={sin}^{\mathrm{2}} {t}\Leftrightarrow \\ $$$${x}=\:\frac{{sin}^{\mathrm{2}} {t}}{\mathrm{1}−{sin}^{\mathrm{2}} {t}}=−\:\frac{\mathrm{1}−{sin}^{\mathrm{2}} {t}\:−\mathrm{1}}{\mathrm{1}−{sin}^{\mathrm{2}} {t}}=−\mathrm{1}\:−\frac{\mathrm{1}}{{sin}^{\mathrm{2}} {t}−\mathrm{1}}\Rightarrow \\ $$$${dx}=\frac{\mathrm{2}{sint}\:{cost}}{\left({sin}^{\mathrm{2}} {t}\:−\mathrm{1}\right)^{\mathrm{2}} }{dt}\:{so} \\ $$$${I}=\:\int\:{t}\:\frac{\mathrm{2}{sint}\:{cost}}{\left(\mathrm{1}−{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}\:=\:\int\:\:\frac{\mathrm{2}{t}\:{sint}\:{cost}}{{cos}^{\mathrm{4}} {t}}{dt} \\ $$$$=\:\int\:\:\:\frac{\mathrm{2}{t}\:{sint}}{{cos}^{\mathrm{3}} {t}}{dt}\:\:{by}\:{parts}\:\:{u}=\mathrm{2}{t}\:{and}\:{v}^{'} =\:\frac{{sint}}{{cos}^{\mathrm{3}} {t}}\:\:\:\:\:\:\:\:\: \\ $$$${I}=\:\:−{t}\:{cos}^{−\mathrm{2}} {t}\:−\int\:\mathrm{2}\left(\frac{−\mathrm{1}}{\mathrm{2}}{cos}^{−\mathrm{2}} {t}\right){dt} \\ $$$$=\:−\frac{{t}}{{cos}^{\mathrm{2}} {t}}\:+\:\int\:\:\frac{{dt}}{{cos}^{\mathrm{2}} {t}}=\:{tant}\:−\frac{{t}}{{cos}^{\mathrm{2}} {t}}\:+{k}\:{but} \\ $$$${cos}^{\mathrm{2}} {t}\:=\mathrm{1}−{sin}^{\mathrm{2}} {t}=\mathrm{1}−\frac{{x}}{{x}+\mathrm{1}}=\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\:{and}\: \\ $$$${tant}=\:\frac{{sint}}{{cost}}=\frac{\sqrt{\frac{{x}}{{x}+\mathrm{1}}}}{\sqrt{\frac{\mathrm{1}}{{x}+\mathrm{1}}}}=\:{x}\:\:\:{so} \\ $$$${I}=\:\sqrt{{x}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}{arcsin}\left(\sqrt{\frac{{x}}{{x}+\mathrm{1}}}\:\right)\:+{k}\:. \\ $$

Commented by abdo imad last updated on 02/Feb/18

the Q is find I= ∫ arcsin((√(x/(x+1))) )dx .

$${the}\:{Q}\:{is}\:{find}\:{I}=\:\int\:{arcsin}\left(\sqrt{\frac{{x}}{{x}+\mathrm{1}}}\:\right){dx}\:. \\ $$

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