Menu Close

calculate-0-1-2-1-x-4-dx-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *