How to integrate ?

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Answer 1 · 2015-12-23T23:28:29

$\int \frac{x^{2}}{\sqrt{4 - x^{2}}} d x$ $intx^2/sqrt(4-x^2)dx$

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

$u = arcsin (\frac{1}{2} x)$ $u = arcsin(1/2x)$

$d x = 2 cos (u) d u$ $dx = 2cos(u)du$

$\int 4 \frac{{sin}^{2} (u)}{2 cos (u)} \cdot 2 cos (u) d u$ $int4sin^2(u)/(2cos(u))*2cos(u)du$ (Just magical)

$\int 4 {sin}^{2} (u) d u$ $int 4sin^2(u)du$

$2 \int 1 - cos (2 u) d u$ $2int1-cos(2u)du$

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

Substitute back

$2 [arcsin (\frac{1}{2} x) - sin (2 arcsin (\frac{1}{2} x))] + C$ $2[arcsin(1/2x)-sin(2arcsin(1/2x))]+C$

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