Question

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

Answer 1

`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)`

Answer 2

`(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`