How do you find the integral of $\frac{x^{2}}{\sqrt{4 x - x^{2}}} d x$ $x^2/sqrt(4x-x^2) dx$ ?

Question

11590 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T08:52:50

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

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

Complete the square in the square root:

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

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

$I = \int {(2 sin θ + 2)}^{2} d θ$ $I=int(2sintheta+2)^2d theta$

Rearrange:

$I = \int (4 {sin}^{2} θ + 8 sin θ + 4) d θ$ $I=int(4sin^2theta+8sintheta+4)d theta$

Apply the identity $cos 2 θ = 1 - 2 {sin}^{2} θ$ $cos2theta=1-2sin^2theta$ :

$I = \int (6 + 8 sin θ - 2 cos 2 θ) d θ$ $I=int(6+8sintheta-2cos2theta)d theta$

Integrate term by term:

$I = 6 θ - 8 cos θ - sin 2 θ + C$ $I=6theta-8costheta-sin2theta+C$

Apply the identity $sin 2 θ = 2 sin θ cos θ$ $sin2theta=2sinthetacostheta$ :

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

Reverse the substitution:

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

How do you find the integral of x2√4x−x2dxx^2/sqrt(4x-x^2) dx?