Question #34e5a

Sep 1, 2017

$\ln \left\mid x + \sqrt{{x}^{2} - 9} \right\mid - \frac{x}{\sqrt{{x}^{2} - 9}} + C$

Explanation:

$I = \int {x}^{2} / {\left({x}^{2} - 9\right)}^{\frac{3}{2}} \mathrm{dx}$

Use the substitution $x = 3 \sec \theta$. This implies that $\mathrm{dx} = 3 \sec \theta \tan \theta d \theta$. Importantly, ${x}^{2} - 9 = 9 {\sec}^{2} \theta - 9 = 9 \left({\sec}^{2} \theta - 1\right) = 9 {\tan}^{2} \theta$.

$I = \int \frac{9 {\sec}^{2} \theta}{9 {\tan}^{2} \theta} ^ \left(\frac{3}{2}\right) \left(3 \sec \theta \tan \theta d \theta\right)$

$I = \int \frac{27 {\sec}^{3} \theta \tan \theta}{27 {\tan}^{3} \theta} d \theta$

$I = \int {\sec}^{3} \frac{\theta}{\tan} ^ 2 \theta d \theta$

$I = \int \frac{1}{\cos} ^ 3 \theta {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$

$I = \int {\csc}^{2} \theta \sec \theta d \theta$

Use ${\csc}^{2} \theta = {\cot}^{2} \theta + 1$:

$I = \int \left({\cot}^{2} \theta + 1\right) \sec \theta d \theta$

$I = \int \left({\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta \frac{1}{\cos} \theta + \sec \theta\right) d \theta$

$I = \int \left(\cot \theta \csc \theta + \sec \theta\right) d \theta$

These have common integrals:

$I = - \csc \theta + \ln \left\mid \sec \theta + \tan \theta \right\mid + C$

Our original substitution was $\sec \theta = \frac{x}{3}$. This means that $\cos \theta = \frac{3}{x}$, so we have a right triangle where the side adjacent to $\theta$ is $3$ and the hypotenuse is $x$. Through the Pythagorean Theorem, the side opposite $\theta$ is $\sqrt{{x}^{2} - 9}$. Then:

• $\tan \theta = \text{opp"/"adj} = \frac{\sqrt{{x}^{2} - 9}}{3}$
• $\csc \theta = \frac{1}{\sin} \theta = \text{hyp"/"opp} = \frac{x}{\sqrt{{x}^{2} - 9}}$

Hence:

$I = \ln \left\mid \frac{x}{3} + \frac{\sqrt{{x}^{2} + 9}}{3} \right\mid - \frac{x}{\sqrt{{x}^{2} - 9}} + C$

Factor $\frac{1}{3}$ from the natural logarithm. Using $\log \left(A B\right) = \log \left(A\right) + \log \left(B\right)$, a $\ln \left(\frac{1}{3}\right)$ will be spit out of the logarithm but it can be neglected because as a constant it will be absorbed into $C$, the constant of integration.

$I = \ln \left\mid x + \sqrt{{x}^{2} - 9} \right\mid - \frac{x}{\sqrt{{x}^{2} - 9}} + C$