Question Number 142886 by ZiYangLee last updated on 06/Jun/21
$$\mathrm{Evaluate}\:\int_{\mathrm{0}} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{4}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:}\:{dx} \\ $$
Answered by Ar Brandon last updated on 06/Jun/21
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{4sin}^{\mathrm{2}} \phi\mathrm{d}\phi=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \left(\mathrm{1}−\mathrm{cos2}\phi\right)\mathrm{d}\phi=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$
Answered by mathmax by abdo last updated on 07/Jun/21
![I=∫_0 ^(1/2) ((4x^2 )/( (√(1−x^2 ))))dx ⇒I=−4 ∫_0 ^(1/2) ((1−x^2 −1)/( (√(1−x^2 ))))dx =−4∫_0 ^(1/2) (√(1−x^2 ))dx+4∫_0 ^(1/2) (dx/( (√(1−x^2 )))) ∫_0 ^(1/2) (√(1−x^2 ))dx =_(x=sinθ) ∫_0 ^(π/6) cos^2 θdθ =∫_0 ^(π/6) ((1+cos(2θ))/2)dθ =(π/(12)) +(1/4)[sin(2θ)]_0 ^(π/6) =(π/(12))+(1/4)(((√3)/2)) =(π/(12))+((√3)/8) ∫_0 ^(1/2) (dx/( (√(1−x^2 ))))=[arcsinx]_0 ^(1/2) =(π/6) ⇒ I=((2π)/3)−4((π/(12))+((√3)/8)) =((2π)/3)−(π/3)−((√3)/2)=(π/3)−((√3)/2)](https://www.tinkutara.com/question/Q142958.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{4x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\:\Rightarrow\mathrm{I}=−\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$=−\mathrm{4}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}+\mathrm{4}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=_{\mathrm{x}=\mathrm{sin}\theta} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\mathrm{d}\theta \\ $$$$=\frac{\pi}{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:=\frac{\pi}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{12}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}=\left[\mathrm{arcsinx}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{6}}\:\Rightarrow \\ $$$$\mathrm{I}=\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{4}\left(\frac{\pi}{\mathrm{12}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\right)\:=\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$