# How do you integrate int dx/(5-4x-x^2)^(5/2) using trig substitutions?

May 1, 2018

$I = \frac{\left(x + 2\right) \left(19 - 8 x - 2 {x}^{2}\right)}{243 \sqrt{5 - 4 x - {x}^{2}}} + C$.

#### Explanation:

Let us note that,

$5 - 4 x - {x}^{2} = 9 - \left(4 + 4 x + {x}^{2}\right) = {3}^{2} - {\left(x + 2\right)}^{2}$.

So, if we subst. $\left(x + 2\right) = 3 \sin y , \text{ then, } \mathrm{dx} = 3 \cos y \mathrm{dy}$.

$\therefore I = \int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right)$,

$= \int \frac{3 \cos y \mathrm{dy}}{9 - 9 {\sin}^{2} y} ^ \left(\frac{5}{2}\right)$,

$= \frac{3}{3} ^ 5 \int \frac{\cos y \mathrm{dy}}{\cos} ^ 5 y$,

$= \frac{1}{81} \int \frac{\mathrm{dy}}{\cos} ^ 4 y$,

$= \frac{1}{81} \int {\sec}^{2} y \cdot {\sec}^{2} y \mathrm{dy}$,

$= \frac{1}{81} \int \left({\tan}^{2} y + 1\right) {\sec}^{2} y \mathrm{dy}$.

Now we subst. $\tan y = t . \therefore {\sec}^{2} y \mathrm{dy} = \mathrm{dt} .$

$\therefore I = \frac{1}{81} \int \left({t}^{2} + 1\right) \mathrm{dt}$,

$= \frac{1}{81} \left({t}^{3} / 3 + t\right)$,

$= \frac{t}{243} \left({t}^{2} + 3\right)$.

Since, $t = \tan y$, we get,

$I = \tan \frac{y}{243} \left({\tan}^{2} y + 3\right)$.

Now, $\frac{x + 2}{3} = \sin y$.

$\therefore \cos y = \sqrt{1 - {\left(x + 2\right)}^{2} / 9} = \frac{1}{3} \sqrt{5 - 4 x - {x}^{2}}$.

$\therefore \tan y = \frac{x + 2}{\sqrt{5 - 4 x - {x}^{2}}}$.

Finally, $I = \frac{1}{243} \cdot \frac{x + 2}{\sqrt{5 - 4 x - {x}^{2}}} \left\{{\left(\frac{x + 2}{\sqrt{5 - 4 x - {x}^{2}}}\right)}^{2} + 3\right\} , i . e . ,$

$I = \frac{\left(x + 2\right) \left(19 - 8 x - 2 {x}^{2}\right)}{243 \sqrt{5 - 4 x - {x}^{2}}} + C$.