How to integrate ?

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Answer 1 · 2015-12-23T23:28:29

$intx^2/sqrt(4-x^2)dx$

Let's $x = 2sin(u)$

$u = arcsin(1/2x)$

$dx = 2cos(u)du$

$int4sin^2(u)/(2cos(u))*2cos(u)du$ (Just magical)

$int 4sin^2(u)du$

$2int1-cos(2u)du$

$[2u-2sin(2u)]+C$

Substitute back

$2[arcsin(1/2x)-sin(2arcsin(1/2x))]+C$

which is $[2arcsin(x/2)-1/2 x sqrt(4-x^2)]$