Question-86407

Question Number 86407 by Power last updated on 28/Mar/20

Commented by mathmax by abdo last updated on 28/Mar/20

A =∫ (√((x−a)(b−x)))dx ⇒A =∫(√(x−a))(√(b−x))dx chsngement (√(x−a))=t give x−a =t^2 ⇒ A =∫ t(√(b−(a+t^2 )))(2t)dt =2 ∫ t^2 (√(b−a−t^2 ))dt =_(t=(√(b−a))sinu) 2 ∫ (b−a)sin^2 u(√(b−a))cosu (√(b−a))cosu du =2(b−a) ∫ sin^2 u cos^2 u du =2(b−a)∫ ((1/2)sin(2u))^2 du =((b−a)/2) ∫ ((1−cos(4u))/2) du =((b−a)/4){ u−(1/4)sin(4u)} +C =((b−a)/4){ u−(1/2)sin(2u)cos(2u)} +C =((b−a)/4){u −sinu cosu (1−2sin^2 u)} +C =((b−a)/2){arcsin((t/( (√(b−a)))))−(t/( (√(b−a))))(√(1−((t/( (√(b−a)))))^2 ))(1−2(t^2 /(b−a)))} +C =((b−a)/2){arcsin(((√(x−a))/( (√(b−a)))))−((√(x−a))/( (√(b−a))))(√(1−(((x−a)/(b−a)))))(1−2((x−a)/(b−a))) } +C

$$\left.{A}\:=\int\:\sqrt{\left({x}−{a}\right)\left({b}−{x}\right.}\right){dx}\:\Rightarrow{A}\:=\int\sqrt{{x}−{a}}\sqrt{{b}−{x}}{dx}\:{chsngement} \\ $$$$\sqrt{{x}−{a}}={t}\:{give}\:{x}−{a}\:={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${A}\:=\int\:\:{t}\sqrt{{b}−\left({a}+{t}^{\mathrm{2}} \right)}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\:{t}^{\mathrm{2}} \sqrt{{b}−{a}−{t}^{\mathrm{2}} }{dt} \\ $$$$=_{{t}=\sqrt{{b}−{a}}{sinu}} \:\:\:\:\mathrm{2}\:\int\:\:\left({b}−{a}\right){sin}^{\mathrm{2}} {u}\sqrt{{b}−{a}}{cosu}\:\sqrt{{b}−{a}}{cosu}\:{du} \\ $$$$=\mathrm{2}\left({b}−{a}\right)\:\int\:{sin}^{\mathrm{2}} {u}\:{cos}^{\mathrm{2}} {u}\:{du} \\ $$$$=\mathrm{2}\left({b}−{a}\right)\int\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{u}\right)\right)^{\mathrm{2}} \:{du} \\ $$$$=\frac{{b}−{a}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{4}{u}\right)}{\mathrm{2}}\:{du} \\ $$$$=\frac{{b}−{a}}{\mathrm{4}}\left\{\:\:{u}−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{4}{u}\right)\right\}\:+{C} \\ $$$$=\frac{{b}−{a}}{\mathrm{4}}\left\{\:{u}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{u}\right){cos}\left(\mathrm{2}{u}\right)\right\}\:+{C} \\ $$$$=\frac{{b}−{a}}{\mathrm{4}}\left\{{u}\:−{sinu}\:{cosu}\:\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {u}\right)\right\}\:+{C} \\ $$$$=\frac{{b}−{a}}{\mathrm{2}}\left\{{arcsin}\left(\frac{{t}}{\:\sqrt{{b}−{a}}}\right)−\frac{{t}}{\:\sqrt{{b}−{a}}}\sqrt{\mathrm{1}−\left(\frac{{t}}{\:\sqrt{{b}−{a}}}\right)^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{2}\frac{{t}^{\mathrm{2}} }{{b}−{a}}\right)\right\}\:+{C} \\ $$$$=\frac{{b}−{a}}{\mathrm{2}}\left\{{arcsin}\left(\frac{\sqrt{{x}−{a}}}{\:\sqrt{{b}−{a}}}\right)−\frac{\sqrt{{x}−{a}}}{\:\sqrt{{b}−{a}}}\sqrt{\mathrm{1}−\left(\frac{{x}−{a}}{{b}−{a}}\right)}\left(\mathrm{1}−\mathrm{2}\frac{{x}−{a}}{{b}−{a}}\right)\:\right\}\:+{C} \\ $$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

(x−a)(b−x) xb−x^2 −ab+ax −x^2 +x(a+b)−ab −{x^2 −2x.((a+b)/2)+(((a+b)/2))^2 +ab−(((a+b)/2))^2 } =(((a+b)/2))^2 −ab−(x−((a+b)/2))^2 =(((a−b)/2))^2 −(x−((a+b)/2))^2 now use formula of ∫(√(A^2 −X^2 )) dX =(X/2)(√(A^2 −X^2 )) +(A^2 /2)sin^(−1) ((X/A))+C ∫(√((((a−b)/2))^2 −(x−((a+b)/2))^2 )) dx =(((x−((a+b)/2)))/2)(√((((a−b)/2))^2 −(x−((a+b)/2))^2 )) +(((((a−b)/2))^2 )/2)sin^(−1) (((x−((a+b)/2))/((a−b)/2)))+C

$$\left({x}−{a}\right)\left({b}−{x}\right) \\ $$$${xb}−{x}^{\mathrm{2}} −{ab}+{ax} \\ $$$$−{x}^{\mathrm{2}} +{x}\left({a}+{b}\right)−{ab} \\ $$$$−\left\{{x}^{\mathrm{2}} −\mathrm{2}{x}.\frac{{a}+{b}}{\mathrm{2}}+\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{ab}−\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \right\} \\ $$$$=\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −{ab}−\left({x}−\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({x}−\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${now}\:{use}\:{formula}\:{of}\:\int\sqrt{{A}^{\mathrm{2}} −{X}^{\mathrm{2}} }\:{dX} \\ $$$$=\frac{{X}}{\mathrm{2}}\sqrt{{A}^{\mathrm{2}} −{X}^{\mathrm{2}} }\:+\frac{{A}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{X}}{{A}}\right)+{C} \\ $$$$\int\sqrt{\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({x}−\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} }\:{dx} \\ $$$$=\frac{\left({x}−\frac{{a}+{b}}{\mathrm{2}}\right)}{\mathrm{2}}\sqrt{\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({x}−\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\frac{\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}−\frac{{a}+{b}}{\mathrm{2}}}{\frac{{a}−{b}}{\mathrm{2}}}\right)+{C} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TANMAY PANACEA. last updated on 28/Mar/20

$${most}\:{welcome}\:{sir} \\ $$

Commented by Power last updated on 28/Mar/20

$$\mathrm{thanks} \\ $$

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