# How do you integrate int (x^2-8x+21)^(3/2) using trig substitutions?

Jun 3, 2018

Use the substitution $x - 4 = \sqrt{5} \tan \theta$ then apply integration by parts.

#### Explanation:

Let

I=int(x^2−8x+21)^(3/2)dx

Complete the square:

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

Apply the substitution $x - 4 = \sqrt{5} \tan \theta$:

$I = 25 \int {\sec}^{5} \theta d \theta$

Apply the integration by parts:

$u \left(\theta\right) = {\sec}^{3} \theta$, $u ' \left(\theta\right) = 3 {\sec}^{3} \theta \tan \theta$
$v ' \left(\theta\right) = {\sec}^{2} \theta$, $v \left(\theta\right) = \tan \theta$

Hence:

$\int {\sec}^{5} \theta d \theta = {\sec}^{3} \theta \tan \theta - 3 \int {\sec}^{3} \theta {\tan}^{2} \theta d \theta$

Rearrange:

$\int {\sec}^{5} \theta d \theta = \frac{1}{4} {\sec}^{3} \theta \tan \theta + \frac{3}{4} \int {\sec}^{3} \theta d \theta$

Similarly:

$\int {\sec}^{3} \theta d \theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln | \sec \theta + \tan \theta | + C$

Hence

$I = 25 \left\{\frac{1}{4} {\sec}^{3} \theta \tan \theta + \frac{3}{8} \sec \theta \tan \theta + \frac{3}{8} \ln | \sec \theta + \tan \theta |\right\} + C$

Reverse the substitution:

$I = \frac{1}{8} \left(x - 4\right) \left(2 {x}^{2} - 16 x + 57\right) \sqrt{{x}^{2} - 8 x + 21} + \frac{75}{8} \ln | x - 4 + \sqrt{{x}^{2} - 8 x + 21} | + C$