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Question Number 10902 by Nayon last updated on 01/Mar/17

                ∫(√(1−x^2 ))dx=?

$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by ridwan balatif last updated on 02/Mar/17

∫(√(1−x^2 ))dx  let: x=sinθ→cosθ=(√(1−x^2 ))         (dx/dθ)=cosθ         dx=cosθ dθ  ∫(√(1−(sinθ)^2 ))(cosθ)dθ  ∫(√(cos^2 θ ))×cosθdθ  ∫cos^2 θdθ  Remember:   cos2θ=−1+2cos^2 θ  cos^2 θ =(1/2)+(1/2)cos2θ !!  ∫cos^2 θdθ=∫((1/2)+(1/2)cos2θ)dθ                        =(1/2)θ+(1/4)sin2θ+C                        =(1/2)θ+(1/2)sinθ×cosθ+C                        =(1/2)sin^(−1) (x)+(1/2)x(√(1−x^2 ))+C

$$\int\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\mathrm{let}:\:\mathrm{x}=\mathrm{sin}\theta\rightarrow\mathrm{cos}\theta=\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{dx}}{\mathrm{d}\theta}=\mathrm{cos}\theta \\ $$$$\:\:\:\:\:\:\:\mathrm{dx}=\mathrm{cos}\theta\:\mathrm{d}\theta \\ $$$$\int\sqrt{\mathrm{1}−\left(\mathrm{sin}\theta\right)^{\mathrm{2}} }\left(\mathrm{cos}\theta\right)\mathrm{d}\theta \\ $$$$\int\sqrt{\mathrm{cos}^{\mathrm{2}} \theta\:}×\mathrm{cos}\theta\mathrm{d}\theta \\ $$$$\int\mathrm{cos}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\mathrm{Remember}:\: \\ $$$$\mathrm{cos2}\theta=−\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \theta \\ $$$$\mathrm{cos}^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos2}\theta\:!! \\ $$$$\int\mathrm{cos}^{\mathrm{2}} \theta\mathrm{d}\theta=\int\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos2}\theta\right)\mathrm{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin2}\theta+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\theta+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\theta×\mathrm{cos}\theta+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$

Commented by Nayon last updated on 01/Mar/17

wrong answer

$${wrong}\:{answer} \\ $$

Commented by sandy_suhendra last updated on 02/Mar/17

what′s the answer?  maybe we should use cosθ=x

$$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{answer}? \\ $$$$\mathrm{maybe}\:\mathrm{we}\:\mathrm{should}\:\mathrm{use}\:\mathrm{cos}\theta=\mathrm{x} \\ $$

Commented by prakash jain last updated on 03/Mar/17

The answer is correct.

$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}.\: \\ $$

Answered by Abdulhalim Ghadady last updated on 02/Mar/17

I=∫(√(1−x^2 ))dx=∫((1−x^2 )/(√(1−x^2 )))dx  =∫(1/(√(1−x^2 )))dx−∫(x^2 /(√(1−x^2 )))dx  =sin^(−1) x−∫((x.x)/(√(1−x^2 )))dx  let:  u=x                 u′=1  v^′ =(x/(√(1−x^2 )))     v=−(√(1−x^2 ))  I=sin^(−1) x−[−x(√(1−x^2 ))−∫−(√(1−x^2 ))dx]  I=sin^(−1) x+x(√(1−x^2 ))−I  2I=sin^(−1) x+x(√(1−x^2 ))  ∴I=(1/2)sin^(−1) x+(1/2)x(√(1−x^2 ))+C

$${I}=\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\int\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}−\int\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\mathrm{sin}^{−\mathrm{1}} {x}−\int\frac{{x}.{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${let}: \\ $$$${u}={x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}'=\mathrm{1} \\ $$$${v}^{'} =\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:\:\:\:{v}=−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${I}=\mathrm{sin}^{−\mathrm{1}} {x}−\left[−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\int−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\right] \\ $$$${I}=\mathrm{sin}^{−\mathrm{1}} {x}+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−{I} \\ $$$$\mathrm{2}{I}=\mathrm{sin}^{−\mathrm{1}} {x}+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\therefore{I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} {x}+\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{C} \\ $$

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