Question

How do you integrate `int (-8x^3)/sqrt(9-x^2)dx` using trigonometric substitution?

Answer 1

Replace `x` with either `3sintheta` or with `3costheta`

Explanation:

`int (-8x^3)/sqrt(9-x^2)dx`

Let `x=3costheta`. This makes

`dx=-3sintheta d theta`, and

`x^3 = 27cos^3 theta`, and

`sqrt(9-x^2) = 3sin theta`

With these substitutions, we get

`int (-8x^3)/sqrt(9-x^2)dx = int (-8(27cos^3theta))/(3sin theta) (-3sintheta) d theta`

`= 8*27intcos^3 theta d theta`

`= 8*27int(1-sin^2 theta) cos theta d theta`

`= 8*27(sin theta-sin^3 theta / 3)+C`

With `cos theta = x/3`, we get `sin theta = sqrt(9-x^2)/3`, so our integral becomes:

`int (-8x^3)/sqrt(9-x^2)dx = 8*27(sqrt(9-x^2)/3 -(sqrt(9-x^2)/3)^3 / 3)`

`= 8*27(sqrt(9-x^2)/3 -((9-x^2)sqrt(9-x^2))/81)+C`

`= 8*27((27sqrt(9-x^2))/81 -((9-x^2)sqrt(9-x^2))/81)+C`

`= 8/3sqrt(9-x^2)(27 -(9-x^2))+C`

`= 8/3sqrt(9-x^2)(x^2+18)+C`