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Question Number 5166 by 1771727373 last updated on 24/Apr/16

∫(√(a^2 −x^2 ))dx = ?

$$\int\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx}\:=\:? \\ $$

Answered by 123456 last updated on 24/Apr/16

x=a sinθ  dx=a cosθdθ  a^2 −x^2 =a^2 sin^2 θ  ∫(√(a^2 cos^2 θ))a cosθdθ    (a cosθ≥0⇒+,a cosθ≤0⇒−)  ±a^2 ∫cos^2 θdθ  ±(a^2 /2)∫1+cos2θdθ  ±(a^2 /2)(θ+(1/2)sinθcosθ)  ±(a^2 /2)sin^(−1) (x/a)+((√(a^2 −x^2 ))/4)+c

$${x}={a}\:\mathrm{sin}\theta \\ $$$${dx}={a}\:\mathrm{cos}\theta{d}\theta \\ $$$${a}^{\mathrm{2}} −{x}^{\mathrm{2}} ={a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta \\ $$$$\int\sqrt{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta}{a}\:\mathrm{cos}\theta{d}\theta\:\:\:\:\left({a}\:\mathrm{cos}\theta\geqslant\mathrm{0}\Rightarrow+,{a}\:\mathrm{cos}\theta\leqslant\mathrm{0}\Rightarrow−\right) \\ $$$$\pm{a}^{\mathrm{2}} \int\mathrm{cos}^{\mathrm{2}} \theta{d}\theta \\ $$$$\pm\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\int\mathrm{1}+\mathrm{cos2}\theta{d}\theta \\ $$$$\pm\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\theta+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\theta\mathrm{cos}\theta\right) \\ $$$$\pm\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{{a}}+\frac{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{4}}+{c} \\ $$

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