Question

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

Answer 1

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

Explanation:

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)`