How do you find the antiderivative of $int x^2/sqrt(4-x^2)dx$ ?

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Answer 1 · 2017-02-22T19:06:09

$2arc sin(x/2)-x/2sqrt(4-x^2)+C.$

We know that,

$(1) : intsqrt(a^2-x^2)dx=x/2sqrt(a^2-x^2)+a^2/2arc sin(x/a)+c_1.$

$(2) : int1/sqrt(a^2-x^2)dx=arc sin(x/a)+c_2.$

Hence, $I=intx^2/sqrt(4-x^2)dx$

$=-int(-x^2)/sqrt(4-x^2)dx=-int{(4-x^2)-4}/sqrt(4-x^2)dx,$

$=-int(4-x^2)/sqrt(4-x^2)dx+4int1/sqrt(4-x^2)dx,$

$=-intsqrt(4-x^2)dx+4arc sin(x/2),............[because, (2)]$

$=-{x/2sqrt(4-x^2)+4/2arc sin(x/2)}+4arc sin (x/2),$

$=2arc sin(x/2)-x/2sqrt(4-x^2)+C.$

Enjoy Maths.!

How do you find the antiderivative of int x^2/sqrt(4-x^2)dxint x^2/sqrt(4-x^2)dx?