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//.](https://www.tinkutara.com/question/Q83839.png)
$$\:\:\int\:\frac{\:\:\mathrm{1}}{\:\sqrt{\mathrm{x}\left(\mathrm{a}−\mathrm{x}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{ax}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\:\int\:\frac{\:\mathrm{1}}{\:\sqrt{−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}.\frac{\mathrm{a}}{\mathrm{2}}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{−\left[\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\right]}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\:\mathrm{1}}{\:\sqrt{\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}\mathrm{dx} \\ $$$$\:=\:\:\mathrm{sin}^{−\mathrm{1}} \frac{\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)}{\frac{\mathrm{a}}{\mathrm{2}}}+\mathrm{C}=\:\mathrm{sin}^{−\mathrm{1}} \frac{\:\:\mathrm{2x}−\mathrm{a}}{\mathrm{a}}\:+\mathrm{C}//. \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 06/Mar/20
$${I}\:=\int\:\:\frac{{dx}}{\:\sqrt{{x}\left({a}−{x}\right)}}\:{vhangement}\:\sqrt{{x}}={t}\:{give}\:{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{\mathrm{2}{tdt}}{{t}\sqrt{{a}−{t}^{\mathrm{2}} }}\:\:=\mathrm{2}\int\:\:\frac{{dt}}{\:\sqrt{{a}−{t}^{\mathrm{2}} }}\:=_{{t}=\sqrt{{a}}{sinu}} \:\mathrm{2}\:\int\:\:\frac{\sqrt{{a}}{cosu}\:{du}}{\:\sqrt{{a}}{cosu}} \\ $$$$=\mathrm{2}{u}\:+{K}\:\:=\mathrm{2}\:{arcsin}\left(\frac{{t}}{\:\sqrt{{a}}}\right)\:+{k}\:=\mathrm{2}{arcsin}\left(\frac{\sqrt{{x}}}{\:\sqrt{{a}}}\right)\:+{K} \\ $$
Answered by TANMAY PANACEA last updated on 06/Mar/20
$${t}^{\mathrm{2}} ={a}−{x} \\ $$$$\int\frac{−\mathrm{2}{tdt}}{\:\sqrt{\left({a}−{t}^{\mathrm{2}} \right)×{t}^{\mathrm{2}} }} \\ $$$$−\mathrm{2}\int\frac{{dt}}{\:\sqrt{{a}−{t}^{\mathrm{2}} }} \\ $$$${t}=\sqrt{{a}}\:×{sin}\alpha \\ $$$$−\mathrm{2}\int\frac{\sqrt{{a}}\:×{cos}\alpha{d}\alpha}{\:\sqrt{{a}}\:×{cos}\alpha}=−\mathrm{2}×\alpha \\ $$$$−\mathrm{2}×{sin}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{{a}}}\right) \\ $$$$−\mathrm{2}×{sin}^{−\mathrm{1}} \left(\sqrt{\frac{{a}−{x}}{{a}}}\:\right)+{c} \\ $$$${pls}\:{check} \\ $$
Commented by Power last updated on 06/Mar/20
$$\mathrm{thanks} \\ $$