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

Dec 19, 2016

I got:

$\int \frac{4 x}{\sqrt{- {x}^{2} - 6 x + 75}} \mathrm{dx}$

$= - 4 \sqrt{- {x}^{2} - 6 x + 75} - 12 \arcsin \left(\frac{x + 3}{\sqrt{84}}\right) + C$

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

$\sqrt{- \left({x}^{2} + 6 x + 9 - 9 - 75\right)}$

$= \sqrt{- \left({\left(x + 3\right)}^{2} - 84\right)} = \sqrt{84 - {\left(x + 3\right)}^{2}}$

Now, this is of the form $\sqrt{{a}^{2} - {x}^{2}}$, so let:

$a = \sqrt{84}$
$x + 3 = a \sin \theta = \sqrt{84} \sin \theta$
$d \left(x + 3\right) = \mathrm{dx} = \sqrt{84} \cos \theta d \theta$
$4 x = 4 \sqrt{84} \sin \theta - 12$

Therefore:

$\implies \sqrt{84 - {\left(x + 3\right)}^{2}} = \sqrt{84 - 84 {\sin}^{2} \theta} = \sqrt{84} \cos \theta$,

and:

$\int \frac{4 x}{\sqrt{- {x}^{2} - 6 x + 75}} \mathrm{dx}$

$= \int \frac{4 \sqrt{84} \sin \theta - 12}{\cancel{\sqrt{84} \cos \theta}} \cancel{\sqrt{84} \cos \theta} d \theta$

$= 4 \sqrt{84} \int \sin \theta d \theta - 12 \int d \theta$

From this point it is quite straightforward to obtain:

$= - 4 \sqrt{84} \cos \theta - 12 \theta$

Almost there. Recall that $x + 3 = \sqrt{84} \sin \theta$ and that $\sqrt{84 - {\left(x + 3\right)}^{2}} = \sqrt{84} \cos \theta$ to get:

$\theta = \arcsin \left(\frac{x + 3}{\sqrt{84}}\right)$
$\cos \theta = \frac{\sqrt{84 - {\left(x + 3\right)}^{2}}}{\sqrt{84}} = \frac{\sqrt{- {x}^{2} - 6 x + 75}}{\sqrt{84}}$

$\implies - 4 \cancel{\sqrt{84}} \frac{\sqrt{- {x}^{2} - 6 x + 75}}{\cancel{\sqrt{84}}} - 12 \arcsin \left(\frac{x + 3}{\sqrt{84}}\right)$
$= \textcolor{b l u e}{- 4 \sqrt{- {x}^{2} - 6 x + 75} - 12 \arcsin \left(\frac{x + 3}{\sqrt{84}}\right) + C}$