Evaluate-0-1-2-4x-2-1-x-2-dx-

Question Number 142886 by ZiYangLee last updated on 06/Jun/21

Evaluate \int_{0}^{\frac{1}{2}} \frac{4 x^{2}}{\sqrt{1 - x^{2}}} d x

Answered by Ar Brandon last updated on 06/Jun/21

= \int_{0}^{\frac{π}{6}} {4 \sin}^{2} ϕ d ϕ = 2 \int_{0}^{\frac{π}{6}} (1 - \cos 2 ϕ) d ϕ = 2 (\frac{π}{6} - \frac{\sqrt{3}}{4})

Answered by mathmax by abdo last updated on 07/Jun/21

I = \int_{0}^{\frac{1}{2}} \frac{{4 x}^{2}}{\sqrt{1 - x^{2}}} dx \Rightarrow I = - 4 \int_{0}^{\frac{1}{2}} \frac{1 - x^{2} - 1}{\sqrt{1 - x^{2}}} dx

= - 4 \int_{0}^{\frac{1}{2}} \sqrt{1 - x^{2}} dx + 4 \int_{0}^{\frac{1}{2}} \frac{dx}{\sqrt{1 - x^{2}}}

\int_{0}^{\frac{1}{2}} \sqrt{1 - x^{2}} dx =_{x = \sin θ} \int_{0}^{\frac{π}{6}} \cos^{2} θ d θ = \int_{0}^{\frac{π}{6}} \frac{1 + \cos (2 θ)}{2} d θ

= \frac{π}{12} + \frac{1}{4} {[\sin (2 θ)]}_{0}^{\frac{π}{6}} = \frac{π}{12} + \frac{1}{4} (\frac{\sqrt{3}}{2}) = \frac{π}{12} + \frac{\sqrt{3}}{8}

\int_{0}^{\frac{1}{2}} \frac{dx}{\sqrt{1 - x^{2}}} = {[arcsinx]}_{0}^{\frac{1}{2}} = \frac{π}{6} \Rightarrow

I = \frac{2 π}{3} - 4 (\frac{π}{12} + \frac{\sqrt{3}}{8}) = \frac{2 π}{3} - \frac{π}{3} - \frac{\sqrt{3}}{2} = \frac{π}{3} - \frac{\sqrt{3}}{2}