To solve this integration, we use the method of integration " Trigonometric Substitution "
Let x=2sinθ , then dx=2cosθdθ
Note that when x=0 then θ=0
and when x=1 , then θ=π/6
Let I=0∫1(4−x2)3x2dx , then I=0∫π/68cos3θ4sin2x∗2cosθdθ=0∫π/6tan2θdθ
Thus I=0∫π/6(sec2θ−1)dθ=[tanθ−θ]0π/6
and then I=[tanθ−θ]0π/6=(1/3)−π/6
Therefore we have that
0∫1(4−x2)3x2dx=(1/3)−(π/6)
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