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

I got:

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

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

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)#