Question

How do you integrate `int x/sqrt(16-x^2)` by trigonometric substitution?

Answer 1

`-sqrt(16-x^2)+C`

Explanation:

Although this is well set up for a shorter, non-trigonometric substitution `(`see: `u=16-x^2)`, we can make the trigonometric substitution `x=4sintheta`. Note that this means that `dx=4costhetad theta`.

Thus:

`intx/sqrt(16-x^2)dx=int(4sintheta)/sqrt(16-16sin^2theta)(4costhetad theta)`

Note that `sqrt(16-16sin^2theta)=4sqrt(1-sin^2theta)=4costheta`.

`=int(4sintheta)/(4costheta)(4costhetad theta)=4intsinthetad theta`

Integrating sine gives:

`=-4costheta+C`

Write this in terms of sine:

`=-4sqrt(1-sin^2theta)+C`

Since `sintheta=x/4`:

`=-4sqrt(1-x^2/16)+C=-4sqrt((16-x^2)/16)+C`

Bringing the `1/16` from the square root as a `1/4`:

`=-sqrt(16-x^2)+C`