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Question Number 51987 by maxmathsup by imad last updated on 01/Jan/19

calculate ∫_0 ^(1/2) (√(1−x^4 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$

Commented by Abdo msup. last updated on 02/Jan/19

let A =∫_0 ^(1/2) (√(1−x^4 ))dx ⇒A=∫_0 ^(1/2) (√(1−x^2 ))(√(1+x^2 ))dx  =_(x^2 =cost)     ∫_1 ^(π/3) (√(1−cost))(√(1+cost))((−sint)/(2(√(cost)))) dt  =−∫_1 ^(π/3)  (√2)sin((t/2))(√2)cos((t/2))  ((sint)/(2(√(cost))))dt  =−∫_1 ^(π/3)  sin((t/2))cos((t/2)) ((sint)/(√(cost)))dt  =− ∫_1 ^(π/3)  (1/2) ((sin^2 t)/(√(cost))) dt =−(1/2) ∫_1 ^(π/3)  ((1−cos^2 t)/(√(cost))) dt  =−(1/2) ∫_1 ^(π/3)  (dt/(√(cost))) +∫_1 ^(π/3)   ((cos^2 t)/(√(cost))) dt ....  be continued...

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\:\Rightarrow{A}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=_{{x}^{\mathrm{2}} ={cost}} \:\:\:\:\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{3}}} \sqrt{\mathrm{1}−{cost}}\sqrt{\mathrm{1}+{cost}}\frac{−{sint}}{\mathrm{2}\sqrt{{cost}}}\:{dt} \\ $$$$=−\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{3}}} \:\sqrt{\mathrm{2}}{sin}\left(\frac{{t}}{\mathrm{2}}\right)\sqrt{\mathrm{2}}{cos}\left(\frac{{t}}{\mathrm{2}}\right)\:\:\frac{{sint}}{\mathrm{2}\sqrt{{cost}}}{dt} \\ $$$$=−\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{3}}} \:{sin}\left(\frac{{t}}{\mathrm{2}}\right){cos}\left(\frac{{t}}{\mathrm{2}}\right)\:\frac{{sint}}{\sqrt{{cost}}}{dt} \\ $$$$=−\:\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{sin}^{\mathrm{2}} {t}}{\sqrt{{cost}}}\:{dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{1}−{cos}^{\mathrm{2}} {t}}{\sqrt{{cost}}}\:{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{dt}}{\sqrt{{cost}}}\:+\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{{cos}^{\mathrm{2}} {t}}{\sqrt{{cost}}}\:{dt}\:.... \\ $$$${be}\:{continued}... \\ $$

Answered by peter frank last updated on 01/Jan/19

∫(√((1+x^2 )(1−x^2 )))  x^2 =sin θ⇒  2xdx=cos θdθ    ∫(√((1+sin θ)(1−sin θ))) dx  ∫(√(1−sin^2  ))   ((cos θdθ)/(2x))    (1/2)∫((cos^2 θdθ)/x)  (1/2)∫((cos^2 θdθ)/(√(sin θ)))  .....

$$\int\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$$${x}^{\mathrm{2}} =\mathrm{sin}\:\theta\Rightarrow\:\:\mathrm{2}{xdx}=\mathrm{cos}\:\theta{d}\theta \\ $$$$ \\ $$$$\int\sqrt{\left(\mathrm{1}+\mathrm{sin}\:\theta\right)\left(\mathrm{1}−\mathrm{sin}\:\theta\right)}\:{dx} \\ $$$$\int\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \:}\:\:\:\frac{\mathrm{cos}\:\theta{d}\theta}{\mathrm{2}{x}}\:\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cos}\:^{\mathrm{2}} \theta{d}\theta}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cos}\:^{\mathrm{2}} \theta{d}\theta}{\sqrt{\mathrm{sin}\:\theta}} \\ $$$$..... \\ $$$$ \\ $$

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