Question Number 47976 by peter frank last updated on 17/Nov/18
Commented by maxmathsup by imad last updated on 17/Nov/18
![let A =∫_(1/4) ^(3/4) ((((√(x−x^2 )))^(−1) )/((dx)^(−2) )) (dx)^(−1) ⇒ A = ∫_(1/4) ^(3/4) (dx/( (√(x−x^2 )))) =∫_(1/4) ^(3/4) (dx/( (√(−(x^2 −2x(1/2) +(1/4)−(1/4)))))) =∫_(1/4) ^(3/4) (dx/( (√((1/(4 ))−(x−(1/2))^2 )))) =_(x−(1/2)=(t/2)) (1/2)∫_(−(1/2)) ^(1/2) ((2dt)/( (√(1−t^2 )))) =[arcsin(t)]_(−(1/2)) ^(1/2) ={(π/6) +(π/6)}=((2π)/6) =(π/3) .](https://www.tinkutara.com/question/Q47977.png)
$${let}\:{A}\:=\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\frac{\mathrm{3}}{\mathrm{4}}} \:\:\:\frac{\left(\sqrt{{x}−{x}^{\mathrm{2}} }\right)^{−\mathrm{1}} }{\left({dx}\right)^{−\mathrm{2}} }\:\left({dx}\right)^{−\mathrm{1}} \:\Rightarrow\:{A}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\frac{\mathrm{3}}{\mathrm{4}}} \:\:\:\frac{{dx}}{\:\sqrt{{x}−{x}^{\mathrm{2}} }} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\frac{\mathrm{3}}{\mathrm{4}}} \:\:\frac{{dx}}{\:\sqrt{−\left({x}^{\mathrm{2}} −\mathrm{2}{x}\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\right)}}\:=\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\frac{\mathrm{3}}{\mathrm{4}}} \:\:\frac{{dx}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}\:}−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{{t}}{\mathrm{2}}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:=\left[{arcsin}\left({t}\right)\right]_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\left\{\frac{\pi}{\mathrm{6}}\:+\frac{\pi}{\mathrm{6}}\right\}=\frac{\mathrm{2}\pi}{\mathrm{6}}\:=\frac{\pi}{\mathrm{3}}\:. \\ $$