# How do you integrate int dx/(5-4x-x^2)^(5/2) by trigonometric substitution?

Mar 22, 2018

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \frac{\left(2 + x\right) \left(19 - 8 x - 2 {x}^{2}\right)}{243 {\left(5 - 4 x - {x}^{2}\right)}^{\frac{3}{2}}}$

#### Explanation:

Complete the square at the denominator:

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \int \frac{\mathrm{dx}}{9 - {\left(2 + x\right)}^{2}} ^ \left(\frac{5}{2}\right)$

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \frac{1}{3} ^ 5 \int \frac{\mathrm{dx}}{1 - {\left(\frac{2 + x}{3}\right)}^{2}} ^ \left(\frac{5}{2}\right)$

Subtitute:

$\frac{2 + x}{3} = \sin t$

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

so that in the interval $\cos t$ is positive and:

$x = - 2 + 3 \sin t$

$\mathrm{dx} = 3 \cos t \mathrm{dt}$

Then:

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \frac{1}{3} ^ 4 \int \frac{\cos t \mathrm{dt}}{1 - {\sin}^{2} t} ^ \left(\frac{5}{2}\right)$

and as:

${\left(1 - {\sin}^{2} t\right)}^{\frac{5}{2}} = {\left(\sqrt{1 - {\sin}^{2} t}\right)}^{5} = {\cos}^{5} t$

we have:

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \frac{1}{3} ^ 4 \int \frac{\mathrm{dt}}{\cos} ^ 4 t = \frac{1}{3} ^ 4 \int {\sec}^{4} t \mathrm{dt}$

Solve the resulting integral using the identity ${\sec}^{2} t = 1 + {\tan}^{2} t$:

$\int {\sec}^{4} t \mathrm{dt} = \int {\sec}^{2} t \cdot {\sec}^{2} t \mathrm{dt}$

$\int {\sec}^{4} t \mathrm{dt} = \int {\sec}^{2} t \left(1 + {\tan}^{2} t\right) \mathrm{dt}$

$\int {\sec}^{4} t \mathrm{dt} = \int {\sec}^{2} t \mathrm{dt} + \int {\tan}^{2} t {\sec}^{2} t \mathrm{dt}$

$\int {\sec}^{4} t \mathrm{dt} = \tan t + {\tan}^{3} \frac{t}{3} + C$

To undo the substitution note that:

$\tan t = \sin \frac{t}{\cos} t$

and as we noted that in the interval $\cos t$ is positive:

$\tan t = \sin \frac{t}{\sqrt{1 - {\sin}^{2} t}}$

so:

tant = ((2+x)/3)/sqrt((1-((2+x)/3)^2)

tant = (2+x)/(sqrt(9-(2+x)^2)

$\tan t = \frac{2 + x}{\sqrt{5 - 4 x - {x}^{2}}}$

Then:

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \frac{1}{3} ^ 4 \left(\frac{2 + x}{\sqrt{5 - 4 x - {x}^{2}}} + {\left(2 + x\right)}^{3} / \left(3 {\left(5 - 4 x - {x}^{2}\right)}^{\frac{3}{2}}\right)\right)$

and simplifying:

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \frac{2 + x}{81 \sqrt{5 - 4 x - {x}^{2}}} \left(1 + {\left(2 + x\right)}^{2} / \left(3 \left(5 - 4 x - {x}^{2}\right)\right)\right)$

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \frac{2 + x}{81 \sqrt{5 - 4 x - {x}^{2}}} \left(\frac{15 - 12 x - 3 {x}^{2} + 4 + 4 x + {x}^{2}}{3 \left(5 - 4 x - {x}^{2}\right)}\right)$

$\int \frac{\mathrm{dx}}{5 - 4 x - {x}^{2}} ^ \left(\frac{5}{2}\right) = \frac{\left(2 + x\right) \left(19 - 8 x - 2 {x}^{2}\right)}{243 {\left(5 - 4 x - {x}^{2}\right)}^{\frac{3}{2}}}$