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Question Number 125837 by joki last updated on 14/Dec/20

prove that :  ∫(√(a^2 −x^2 ))dx=_ (x/2)(√(a^2 −x^2 ))+(a^2 /2)sin^(−1) ((x/a))+c

$${prove}\:{that}\:: \\ $$$$\int\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx}=_{} \frac{{x}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)+{c} \\ $$

Answered by Dwaipayan Shikari last updated on 14/Dec/20

Take  x=asinθ   ,  1=acosθ(dθ/dx)  ∫(√(a^2 −x^2 )) dx = ∫acosθ(√(a^2 −a^2 sin^2 θ))  dθ  =a^2 ∫cos^2 θ dθ =(a^2 /2)∫1+cos2θ dθ  =((a^2 θ)/2)+(a^2 /4)sin2θ +C  θ=sin^(−1) (x/a)    sin2θ =2(x/a^2 )(√(a−x^2 ))  So ∫(√(a^2 −x^2 )) dx = ((a^2 sin^(−1) (x/a))/2)+(x/2)(√(a^2 −x^2 )) +C

$${Take}\:\:{x}={asin}\theta\:\:\:,\:\:\mathrm{1}={acos}\theta\frac{{d}\theta}{{dx}} \\ $$$$\int\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx}\:=\:\int{acos}\theta\sqrt{{a}^{\mathrm{2}} −{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\:\:{d}\theta \\ $$$$={a}^{\mathrm{2}} \int{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\int\mathrm{1}+{cos}\mathrm{2}\theta\:{d}\theta\:\:=\frac{{a}^{\mathrm{2}} \theta}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}{sin}\mathrm{2}\theta\:+{C} \\ $$$$\theta={sin}^{−\mathrm{1}} \frac{{x}}{{a}}\:\:\:\:{sin}\mathrm{2}\theta\:=\mathrm{2}\frac{{x}}{{a}^{\mathrm{2}} }\sqrt{{a}−{x}^{\mathrm{2}} } \\ $$$${So}\:\int\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx}\:=\:\frac{{a}^{\mathrm{2}} {sin}^{−\mathrm{1}} \frac{{x}}{{a}}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:+{C} \\ $$

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