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Question Number 101531 by mhmd last updated on 03/Jul/20
find ∫(√(ax−x^2 ))dx
$${find}\:\int\sqrt{{ax}−{x}^{\mathrm{2}} }{dx} \\ $$
Answered by Mr.D.N. last updated on 03/Jul/20
  I = ∫(√(ax−x^2 ))dx     = ∫(√(−(x^2 −ax)))   dx    =  ∫  (√(((a/2))^2 −(x−(a/2))^2 ))     dx    =  (((x−(a/2))(√(((a/2))^2 −(x−(a/2))^2 )))/2) + ((((a/2))^2 )/2)sin^(−1) (((x−(a/2)))/(a/2))+C  = ((((2x−a)/2)(√(ax−x^2 )))/2) + (a^2 /8) sin^(−1) ((2x−a)/a)+C  = (((2x−a)(√(ax−x^2 )))/4) + (a^2 /8) sin^(−1)  (((2x−a))/a) +C//.
$$\:\:\mathrm{I}\:=\:\int\sqrt{\mathrm{ax}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\:=\:\int\sqrt{−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{ax}\right)}\:\:\:\mathrm{dx} \\ $$$$\:\:=\:\:\int\:\:\sqrt{\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:\:\:\:\mathrm{dx} \\ $$$$\:\:=\:\:\frac{\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)\sqrt{\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}{\mathrm{2}}\:+\:\frac{\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)}{\frac{\mathrm{a}}{\mathrm{2}}}+\mathrm{C} \\ $$$$=\:\frac{\frac{\mathrm{2x}−\mathrm{a}}{\mathrm{2}}\sqrt{\mathrm{ax}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{2}}\:+\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{8}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2x}−\mathrm{a}}{\mathrm{a}}+\mathrm{C} \\ $$$$=\:\frac{\left(\mathrm{2x}−\mathrm{a}\right)\sqrt{\mathrm{ax}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{4}}\:+\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{8}}\:\mathrm{sin}^{−\mathrm{1}} \:\frac{\left(\mathrm{2x}−\mathrm{a}\right)}{\mathrm{a}}\:+\mathrm{C}//. \\ $$
Commented by mhmd last updated on 03/Jul/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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