To solve this integration, we use the method of integration " Trigonometric Substitution "
Let x = 2 s i n θ x=2sin \theta x = 2 s in θ , then d x = 2 c o s θ d θ dx=2cos \theta d\theta d x = 2 cos θ d θ
Note that when x = 0 x=0 x = 0 then θ = 0 \theta =0 θ = 0
and when x = 1 x=1 x = 1 , then θ = π / 6 \theta=\pi/6 θ = π /6
Let I = ∫ 0 1 x 2 ( 4 − x 2 ) 3 d x I=\int\limits_{0}^{1} \frac { x^{2}} {(\sqrt4-x^{2} )^{3}} dx I = 0 ∫ 1 ( 4 − x 2 ) 3 x 2 d x , then I = ∫ 0 π / 6 4 s i n 2 x 8 c o s 3 θ ∗ 2 c o s θ d θ = ∫ 0 π / 6 t a n 2 θ d θ I=\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 I = 0 ∫ π /6 8 co s 3 θ 4 s i n 2 x ∗ 2 cos θ d θ = 0 ∫ π /6 t a n 2 θ d θ
Thus I = ∫ 0 π / 6 ( s e c 2 θ − 1 ) d θ = [ t a n θ − θ ] 0 π / 6 I=\int\limits_{0}^{\pi/6} (sec^{2}\theta-1)d\theta=[tan\theta-\theta]^{\pi/6}_{0} I = 0 ∫ π /6 ( se c 2 θ − 1 ) d θ = [ t an θ − θ ] 0 π /6
and then I = [ t a n θ − θ ] 0 π / 6 = ( 1 / 3 ) − π / 6 I=[tan\theta-\theta]^{\pi/6}_{0}=(1/\sqrt3)-\pi/6 I = [ t an θ − θ ] 0 π /6 = ( 1/ 3 ) − π /6
Therefore we have that
∫ 0 1 x 2 ( 4 − x 2 ) 3 d x = ( 1 / 3 ) − ( π / 6 ) \int\limits_{0}^{1} \frac { x^{2}} {(\sqrt4-x^{2} )^{3}} dx=(1/\sqrt3)-(\pi/6) 0 ∫ 1 ( 4 − x 2 ) 3 x 2 d x = ( 1/ 3 ) − ( π /6 )
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