# How do you integrate int x^3 sqrt(-x^2 - 16x-39)dx using trigonometric substitution?

Jan 25, 2017

$\int {x}^{3} \sqrt{- {x}^{2} - 16 x - 39} \mathrm{dx} = 25 \int {\left(5 \sin t + 8\right)}^{3} \left(1 - {\sin}^{2} t\right) \mathrm{dt}$

#### Explanation:

Write:

$- {x}^{2} - 16 x - 39 = - {x}^{2} - 16 x - 64 + 25 = 25 - {\left(x - 8\right)}^{2}$

so that we get a difference of squares under the root.

Now substitute:

$x - 8 = 5 \sin t$

$\mathrm{dx} = 5 \cos t \mathrm{dt}$

As the function is defined only for:

$\left\mid x - 8 \right\mid \le 5$

we have $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

Performing the substitution we get:

$\int {x}^{3} \sqrt{- {x}^{2} - 16 x - 39} \mathrm{dx} = 5 \int {\left(5 \sin t + 8\right)}^{3} \sqrt{25 - 25 {\sin}^{2} t} \cos t \mathrm{dt} = 25 \int {\left(5 \sin t + 8\right)}^{3} {\cos}^{2} t \mathrm{dt} = 25 \int {\left(5 \sin t + 8\right)}^{3} \left(1 - {\sin}^{2} t\right) \mathrm{dt}$

where in $\cos t = \pm \sqrt{1 - {\sin}^{2} t}$ we took the plus sign as for $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ the cosine is positive.

We have now to integrate a polynomial in $\sin t$, which can be easily calculated using reduction formulas.