Question #124047
Evaluate the integral ∫¹ₒ [x²⁄(√4-x²)ᶟ]dx
1
Expert's answer
2020-06-29T19:05:27-0400

To solve this integration, we use the method of integration " Trigonometric Substitution "

Let x=2sinθx=2sin \theta , then dx=2cosθdθdx=2cos \theta d\theta

Note that when x=0x=0 then θ=0\theta =0

and when x=1x=1 , then θ=π/6\theta=\pi/6


Let I=01x2(4x2)3dxI=\int\limits_{0}^{1} \frac { x^{2}} {(\sqrt4-x^{2} )^{3}} dx , then I=0π/64sin2x8cos3θ2cosθdθ=0π/6tan2θdθI=\int\limits_{0}^{\pi/6} \frac { 4sin^{2}x} {8cos^{3}\theta} *2cos \theta d\theta =\int\limits_{0}^{\pi/6} tan^{2}\theta d\theta


Thus I=0π/6(sec2θ1)dθ=[tanθθ]0π/6I=\int\limits_{0}^{\pi/6} (sec^{2}\theta-1)d\theta=[tan\theta-\theta]^{\pi/6}_{0}

and then I=[tanθθ]0π/6=(1/3)π/6I=[tan\theta-\theta]^{\pi/6}_{0}=(1/\sqrt3)-\pi/6


Therefore we have that

01x2(4x2)3dx=(1/3)(π/6)\int\limits_{0}^{1} \frac { x^{2}} {(\sqrt4-x^{2} )^{3}} dx=(1/\sqrt3)-(\pi/6)







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