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Question Number 83834 by Power last updated on 06/Mar/20

Commented by niroj last updated on 06/Mar/20

∫ (( 1)/(√(x(a−x))))dx = ∫ (1/(√(ax−x^2 )))dx= ∫ (( 1)/(√(−(x^2 −2x.(a/2)+(a^2 /4)−(a^2 /4)))))dx = ∫ (1/(√(−[(x−(a/2))^2 −(a^2 /4)])))dx = ∫ (( 1)/(√(((a/2))^2 −(x−(a/2))^2 )))dx = sin^(−1) (((x−(a/2)))/(a/2))+C= sin^(−1) (( 2x−a)/a) +C//.

$$\:\:\int\:\frac{\:\:\mathrm{1}}{\sqrt{\mathrm{x}\left(\mathrm{a}−\mathrm{x}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\sqrt{\mathrm{ax}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\:\int\:\frac{\:\mathrm{1}}{\sqrt{−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}.\frac{\mathrm{a}}{\mathrm{2}}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\sqrt{−\left[\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\right]}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\:\mathrm{1}}{\sqrt{\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}\mathrm{dx} \\ $$$$\:=\:\:\mathrm{sin}^{−\mathrm{1}} \frac{\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)}{\frac{\mathrm{a}}{\mathrm{2}}}+\mathrm{C}=\:\mathrm{sin}^{−\mathrm{1}} \frac{\:\:\mathrm{2x}−\mathrm{a}}{\mathrm{a}}\:+\mathrm{C}//. \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 06/Mar/20

I =∫ (dx/(√(x(a−x)))) vhangement (√x)=t give x=t^2 ⇒ I =∫ ((2tdt)/(t(√(a−t^2 )))) =2∫ (dt/(√(a−t^2 ))) =_(t=(√a)sinu) 2 ∫ (((√a)cosu du)/((√a)cosu)) =2u +K =2 arcsin((t/(√a))) +k =2arcsin(((√x)/(√a))) +K

$${I}\:=\int\:\:\frac{{dx}}{\sqrt{{x}\left({a}−{x}\right)}}\:{vhangement}\:\sqrt{{x}}={t}\:{give}\:{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{\mathrm{2}{tdt}}{{t}\sqrt{{a}−{t}^{\mathrm{2}} }}\:\:=\mathrm{2}\int\:\:\frac{{dt}}{\sqrt{{a}−{t}^{\mathrm{2}} }}\:=_{{t}=\sqrt{{a}}{sinu}} \:\mathrm{2}\:\int\:\:\frac{\sqrt{{a}}{cosu}\:{du}}{\sqrt{{a}}{cosu}} \\ $$$$=\mathrm{2}{u}\:+{K}\:\:=\mathrm{2}\:{arcsin}\left(\frac{{t}}{\sqrt{{a}}}\right)\:+{k}\:=\mathrm{2}{arcsin}\left(\frac{\sqrt{{x}}}{\sqrt{{a}}}\right)\:+{K} \\ $$

Answered by TANMAY PANACEA last updated on 06/Mar/20

t^2 =a−x ∫((−2tdt)/(√((a−t^2 )×t^2 ))) −2∫(dt/(√(a−t^2 ))) t=(√a) ×sinα −2∫(((√a) ×cosαdα)/((√a) ×cosα))=−2×α −2×sin^(−1) ((t/(√a))) −2×sin^(−1) ((√((a−x)/a)) )+c pls check

$${t}^{\mathrm{2}} ={a}−{x} \\ $$$$\int\frac{−\mathrm{2}{tdt}}{\sqrt{\left({a}−{t}^{\mathrm{2}} \right)×{t}^{\mathrm{2}} }} \\ $$$$−\mathrm{2}\int\frac{{dt}}{\sqrt{{a}−{t}^{\mathrm{2}} }} \\ $$$${t}=\sqrt{{a}}\:×{sin}\alpha \\ $$$$−\mathrm{2}\int\frac{\sqrt{{a}}\:×{cos}\alpha{d}\alpha}{\sqrt{{a}}\:×{cos}\alpha}=−\mathrm{2}×\alpha \\ $$$$−\mathrm{2}×{sin}^{−\mathrm{1}} \left(\frac{{t}}{\sqrt{{a}}}\right) \\ $$$$−\mathrm{2}×{sin}^{−\mathrm{1}} \left(\sqrt{\frac{{a}−{x}}{{a}}}\:\right)+{c} \\ $$$${pls}\:{check} \\ $$

Commented by Power last updated on 06/Mar/20

thanks

$$\mathrm{thanks} \\ $$

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