How do you integrate int (x^2 ) / sqrt(9 - x^2) dx using trigonometric substitution?

2 Answers
Apr 1, 2018

int(x^2/sqrt(9-x^2))dx=9/2arcsin(x/3)-1/2xsqrt(9-x^2)

Explanation:

Let

x=3sintheta

x^2=9sin^2theta

dx=3costhetad theta

We then have

int(27sin^2thetacostheta)/(sqrt(9(1-sin^2theta))d theta

Recall that 1-sin^2theta=cos^2theta. Apply the identity:

9int(sin^2thetacostheta)/sqrt(cos^2theta)d theta

9int(sin^2thetacancelcostheta)/(cancelcostheta)d theta

9intsin^2thetad theta

Recall the identity sin^2theta=1/2(1-cos2theta):

9/2int(1-cos2theta)d theta

Integrate:

9/2int(1-cos2theta)d theta=9/2theta-9/4sin2theta+C

We need to revert back to x. Recalling that x=3sintheta, sintheta=x/3, theta=arcsin(x/3)

sin2theta is still needed. Recalling that sin2theta=2sinthetacostheta, sin^2theta+cos^2theta=1:

x^2/9+cos^2theta=9/9
cos^2theta=(9-x^2)/9
costheta=sqrt(9-x^2)/3

Then sin2theta=2sinthetacostheta=2(x/3)(sqrt(9-x^2)/3)=(2xsqrt(9-x^2))/9

Then,

int(x^2/sqrt(9-x^2))dx=9/2arcsin(x/3)-1/2xsqrt(9-x^2)

Apr 1, 2018

(9/2)(arcsin(x/3)-1/2sin(2arcsin(x/3))+C

Explanation:

intx^2/(sqrt(9-x^2)) dx= x^2/(sqrt(3^2-x^2))dx using our trig identities for integrals let us substitute 3sinphi = x and dx = 3cosphi dphi

therefore 3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9-(9sin^2phi))

3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9(1-(sin^2phi))
where 1-sin^2phi = cos^2phi

3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9(cos^2phi))

3int(9sin^2phi)(cancelcosphi)(dphi)/(3(cancelcosphi))

cancel3int(9sin^2phi)(dphi)/(cancel3)

int(9sin^2phi)(dphi) = 9int(sin^2phi)(dphi)

Use the identity from trig where cos(2phi) = 1-2sin^2phi rearranging to solve such that (cos(2phi) -1)/(-2)= sin^2phi or (1-cos(2phi))/(2)= sin^2phi

9int(1-cos(2phi))/(2)(dphi) = 9/2int(1-cos(2phi))(dphi) use subtraction rules of the integral

9/2int(dphi)-9/2intcos(2phi)dphi

9/2int(dphi) = (9/2)phi

-9/2intcos(2phi)dphi = -9/4sin(2phi)

Now piece together and substitute in

(9/2)phi-9/4sin(2phi)+C

phi = arcsin(x/3)

(9/2)phi-9/4sin(2phi)+C

(9/2)(arcsin(x/3)-1/2sin(2arcsin(x/3))+C