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

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Answer 1 · 2018-03-08T08:52:50

Use the substitution $x-2=2sintheta$ .

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

Complete the square in the square root:

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

Apply the substitution $x-2=2sintheta$ :

$I=int(2sintheta+2)^2d theta$

Rearrange:

$I=int(4sin^2theta+8sintheta+4)d theta$

Apply the identity $cos2theta=1-2sin^2theta$ :

$I=int(6+8sintheta-2cos2theta)d theta$

Integrate term by term:

$I=6theta-8costheta-sin2theta+C$

Apply the identity $sin2theta=2sinthetacostheta$ :

$I=6theta-8costheta-2sinthetacostheta+C$

Reverse the substitution:

$I=6sin^(-1)((x-2)/2)-1/2(x+6)sqrt(4x-x^2)$

How do you find the integral of x^2/sqrt(4x-x^2) dxx^2/sqrt(4x-x^2) dx?