Question

Evaluate the following integrals: $$ \int \frac{x^{2} d x}{\sqrt{x^{2}-16}} $$

Short answer

Question: Evaluate the integral: $$\int\frac{x^2}{\sqrt{x^2-16}}dx$$ Answer: $$4\sqrt{x^2-16}+C$$

Step-by-step solution

Perform Trigonometric Substitution

Let's start the trigonometric substitution by letting: $$ x = 4\sec(\theta) $$ From this substitution, the derivative of x with respect to θ is: $$ dx = 4\sec(\theta)\tan(\theta)d\theta $$ Now, let's adjust the expression inside the square root: $$ \sqrt{x^2 - 16} = \sqrt{(4\sec(\theta))^2 - 16} $$ Simplify the expression: $$ \sqrt{16\sec^2(\theta) - 16} = 4\sqrt{\sec^2(\theta) - 1} $$ Since \(\sec^2(\theta) - 1 = \tan^2(\theta)\), then: $$ \sqrt{x^2 - 16} = 4\tan(\theta). $$

Rewrite Integral in Terms of Substituted Variable

Now let's replace the expression with the substitution: $$ \int \frac{x^2}{\sqrt{x^2-16}}dx = \int \frac{(4\sec(\theta))^2}{4\tan(\theta)}(4\sec(\theta)\tan(\theta)d\theta) $$ Simplify the integral: $$ \int 16\sec^2(\theta)d\theta $$

Compute Integral

Compute the integral of the simplified expression: $$ \int 16\sec^2(\theta)d\theta = 16\int\sec^2(\theta)d\theta $$ The integral of \(\sec^2(\theta)\) is \(\tan(\theta)\), so: $$ 16\int\sec^2(\theta)d\theta = 16\tan(\theta) + C $$ Where C is the constant of integration.

Transform back to Original Variable

Now, we'll replace θ with the original variable x. Recall our original substitution: $$ x = 4\sec(\theta) $$ Now, we'll use the fact that \(\sec^2(\theta) - 1 = \tan^2(\theta)\): $$ \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \frac{\sqrt{x^2 -16}}{4} $$ Thus, the final answer is: $$ 16\tan(\theta) + C = 16\left(\frac{\sqrt{x^2 -16}}{4}\right) + C = 4\sqrt{x^2 -16} + C $$