Question

How do you integrate `int (4x)/sqrt(-x^2-6x+75)dx` using trigonometric substitution?

Answer 1

I got:

`int (4x)/(sqrt(-x^2 - 6x + 75))dx`

`= -4sqrt(-x^2 - 6x + 75) - 12arcsin((x+3)/sqrt84) + C`

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Usually by completing the square. Notice how we can add and subtract `9` within `sqrt(-(x^2 + 6x - 75))` like so:

`sqrt(-(x^2 + 6x + 9 - 9 - 75))`

`= sqrt(-((x+3)^2 - 84)) = sqrt(84 - (x+3)^2)`

Now, this is of the form `sqrt(a^2 - x^2)`, so let:

`a = sqrt84`
`x + 3 = asintheta = sqrt84 sintheta`
`d(x+3) = dx = sqrt84 costhetad theta`
`4x = 4sqrt84 sintheta - 12`

Therefore:

`=> sqrt(84 - (x+3)^2) = sqrt(84 - 84sin^2theta) = sqrt84 costheta`,

and:

`int (4x)/(sqrt(-x^2 - 6x +75))dx`

`= int(4sqrt84 sintheta - 12)/cancel(sqrt84 costheta)cancel(sqrt84 costheta)d theta`

`= 4sqrt84 int sinthetad theta - 12int d theta`

From this point it is quite straightforward to obtain:

`= -4sqrt84 costheta - 12theta`

Almost there. Recall that `x + 3 = sqrt84 sintheta` and that `sqrt(84 - (x+3)^2) = sqrt84 costheta` to get:

`theta = arcsin((x+3)/sqrt84)`
`costheta = sqrt(84 - (x+3)^2)/sqrt84 = sqrt(-x^2 - 6x + 75)/sqrt84`

Therefore, our answer becomes:

`=> -4cancel(sqrt84)(sqrt(-x^2 - 6x + 75))/cancel(sqrt84) - 12arcsin((x+3)/sqrt84)`

`= color(blue)(-4sqrt(-x^2 - 6x + 75) - 12arcsin((x+3)/sqrt84) + C)`