# How do you integrate int x^3/sqrt(4x^2+8x+82) dx using trigonometric substitution?

Jun 5, 2018

Use the substitution $x + 1 = \frac{\sqrt{78}}{2} \tan \theta$.

#### Explanation:

Let

$I = \int {x}^{3} / \sqrt{4 {x}^{2} + 8 x + 82} \mathrm{dx}$

Complete the square in the denominator:

$I = \int {x}^{3} / \sqrt{4 {\left(x + 1\right)}^{2} + 78} \mathrm{dx}$

Apply the substitution $x + 1 = \frac{\sqrt{78}}{2} \tan \theta$:

$I = \int {\left(\frac{\sqrt{78}}{2} \tan \theta - 1\right)}^{3} / \left(\sqrt{78} \sec \theta\right) \left(\frac{\sqrt{78}}{2} {\sec}^{2} \theta d \theta\right)$

Let $a = \frac{\sqrt{78}}{2}$:

$I = \frac{1}{2} \int \left({a}^{3} {\tan}^{3} \theta - 3 {a}^{2} {\tan}^{2} \theta + 3 a \tan \theta - 1\right) \sec \theta d \theta$

Since ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$:

$I = \frac{1}{2} \int \left\{{a}^{3} {\sec}^{3} \theta \tan \theta + \left(3 a - {a}^{3}\right) \sec \theta \tan \theta - 3 {a}^{2} {\sec}^{3} \theta + \left(3 {a}^{2} - 1\right) \sec \theta\right\} d \theta$

These are all known integrals:

$I = \frac{1}{2} \left\{\frac{1}{3} {a}^{3} {\sec}^{3} \theta + \left(3 a - {a}^{3}\right) \sec \theta - \frac{3}{2} {a}^{2} \left(\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right) + \left(3 {a}^{2} - 1\right) \ln | \sec \theta + \tan \theta |\right\} + C$

Rearrange and rescale C:

$I = \frac{1}{12} \left(2 {\left(a \tan \theta\right)}^{2} - 9 \left(a \tan \theta\right) + 18 - 4 {a}^{2}\right) \left(a \sec \theta\right) + \frac{1}{8} \left(6 {a}^{2} - 4\right) \ln | a \tan \theta + a \sec \theta | + C$

Reverse the substitutions:

$I = \frac{1}{12} \left(2 {x}^{2} - 5 x - 67\right) \sqrt{{\left(x + 1\right)}^{2} + \frac{39}{2}} + \frac{113}{8} \ln | \left(x + 1\right) + \sqrt{{\left(x + 1\right)}^{2} + \frac{39}{2}} | + C$