How do you integrate int ( x^2 / sqrt(4 - x^2) ) dx(x24x2)dx?

1 Answer
Aug 12, 2018

int(x^2)/sqrt(4-x^2)dx=2sin^(-1)(x/2)-x/2sqrt(4-x^2)+Cx24x2dx=2sin1(x2)x24x2+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 !