How do you find the integral of (x^2)/(16-x^2)^(1/2)?

1 Answer
Apr 20, 2018

intx^2/(16-x^2)^(1/2)dx=8arcsin(x/4)-(xsqrt(16-x^2))/2+C

Explanation:

Rewrite with a root:

intx^2/sqrt(16-x^2)dx

For integrals involving the root sqrt(a^2-x^2), we use the substitution x=asintheta.

Here, a^2=16, a=4, x=4sintheta, dx=4costhetad theta and we get

int(16sin^2theta4costhetad theta)/sqrt(16(1-sin^2theta))

Recalling that 1-sin^2theta=cos^2theta,we get

=16int(sin^2thetacosthetad theta)/sqrt(cos^2theta)

=16int(sin^2thetacancelcostheta)/(cancelcostheta)d theta

=16intsin^2thetad theta

Recalling that sin^2theta=1/2(1-cos2theta), we get

16/2int(1-cos2thetad theta)=8(theta-1/2sin2theta)+C

We need to get things in terms of x.

Recalling that x=4sintheta, sintheta=x/4, theta=arcsin(x/4)

To determine 1/2sin2theta, recall the identity 1/2sin2theta=sinthetacostheta. To use this, we need to determine the cosine using the below Pythagorean identity:

sin^2theta+cos^2theta=1
cos^2theta=1-sin^2theta
cos^2theta=1-x^2/16

costheta=sqrt(16-x^2)/4

Thus, 1/2sin2theta=(x/4)sqrt(16-x^2)/4=(xsqrt(16-x^2))/16

And

intx^2/(16-x^2)^(1/2)dx=8arcsin(x/4)-(xsqrt(16-x^2))/2+C