Question

How do you integrate `int ( x^2 / sqrt(4 - x^2) ) dx`?

Answer 1

`int(x^2)/sqrt(4-x^2)dx=2sin^(-1)(x/2)-x/2sqrt(4-x^2)+C`, `C in RR`

Explanation:

`I=int(x^2)/sqrt(4-x^2)dx`
Let `x=2sin(theta)`

`dx=2cos(theta)d theta`

So:

`I=int(4sin(theta)^2)/sqrt(4-4sin(theta)^2)*2cos(theta)d theta`

`=2int(2sin(theta)^2cos(theta))/sqrt(1-sin(theta)^2)d theta`

Because `1-sin(theta)^2=cos(theta)^2`,

`I=2int2sin(theta)^2d theta`

Because `2sin(theta)^2=1-cos(2theta)`

`I=2int(1-cos(2theta))d theta`

`=2theta-sin(2theta)`

`=2(theta-sin(theta)cos(theta))`

Finally, because `theta=sin^(-1)(x/2)`, and `cos(sin^(-1)(x))=sqrt(1-x^2)`

`I=2sin^(-1)(x/2)-xsqrt(1-x^2/4)`

`=2sin^(-1)(x/2)-x/2sqrt(4-x^2)+C`, `C in RR`

\0/ Here's our answer !