Question

How do you integrate `int x^3/sqrt(144-x^2)dx` using trigonometric substitution?

Answer 1

`-1/3sqrt(144-x^2)(x^2+288)+C`

Explanation:

Use the substitution `x=12sintheta`. From here we see that `dx=12costhetad theta`. Substituting:

`intx^3/sqrt(144-x^2)dx=int(1728sin^3theta(12costheta))/sqrt(144-144sin^2theta)d theta`

Factoring `sqrt(1/144)=1/12` from the square root:

`=int(1728sin^3theta(12costheta))/(12sqrt(1-sin^2theta))d theta`

Canceling the `12`s and recalling that since `sin^2theta+cos^2theta=1`, we know that `costheta=sqrt(1-sin^2theta)`:

`=1728int(sin^3thetacostheta)/costhetad theta=1728intsin^3thetad theta`

We will use here `sin^2theta=1-cos^2theta`:

`=1728intsin^2thetasinthetad theta=1728int(1-cos^2theta)sinthetad theta`

Now, using the substitution `u=costheta`, so that `du=-sinthetad theta`:

`=-1728int(1-u^2)du=1728intu^2du-1728intdu`

Integrating using the power rule for integration:

`=1728(u^3/3)-1728u=576u^3-1728u`

Reverse substituting with `u=costheta`:

`=576cos^3theta-1728costheta`

Because our substitution is `sintheta=x/12`, we should write this in terms of sine functions. That is, `costheta=sqrt(1-sin^2theta)` and `cos^3theta=(1-sin^2theta)^(3/2)`:

`=576(1-sin^2theta)^(3/2)-1728sqrt(1-sin^2theta)`

Factoring:

`=sqrt(1-sin^2theta)(576(1-sin^2theta)-1728)`

Now using `sintheta=x/12`:

`=sqrt(1-x^2/144)(576(1-x^2/144)-1728)`

`=sqrt((144-x^2)/144)(576-4x^2-1728)`

Factoring `sqrt(1/144)=1/12` from the square root and `-4` from the parentheses:

`=-1/3sqrt(144-x^2)(x^2+288)+C`