Question

Trigonometric substitutions Evaluate the following integrals using trigonometric substitution. $$\int \frac{\sqrt{9 x^{2}-25}}{x^{3}} d x, x > \frac{5}{3}$$

Short answer

Question: Find the value of the indefinite integral $\int \frac{\sqrt{9x^2 - 25}}{x^3} dx$ for $x > \frac{5}{3}$. Answer: The indefinite integral is given by the expression: $-\frac{25}{27} (\frac{3}{x} \frac{\sqrt{25x^2 - 9}}{5} + \theta) + C$.

Step-by-step solution

Make the substitution

Substitute \(x = \frac{5}{3} \sec(\theta)\) in the given integral: $$\int \frac{\sqrt{9(\frac{5}{3} \sec(\theta))^2 - 25}}{(\frac{5}{3} \sec(\theta))^3} dx$$

Find the derivative of x with respect to θ

Differentiate the expression \(x = \frac{5}{3} \sec(\theta)\) with respect to \(\theta\): $$\frac{d x}{d \theta} = \frac{5}{3} \sec(\theta) \tan(\theta)$$

Determine the relation between dx and dθ

Solve for \(dx\): $$dx = \frac{5}{3} \sec(\theta) \tan(\theta) d \theta$$

Simplify the expression inside the square root

Simplify the expression inside the square root: $$\sqrt{9(\frac{5}{3} \sec(\theta))^2 - 25} = \sqrt{25 \sec^2(\theta) - 25} = 5\sqrt{\sec^2(\theta) - 1} = 5 \ |\tan(\theta)|$$ Since \(x > \frac{5}{3}\), we know that \(\theta > 0\), so we get: $$5\sqrt{\sec^2(\theta) - 1} = 5 \tan(\theta)$$

Substitute and integrate

Now substitute the expressions for x, \(\sqrt{9x^2 - 25}\), and dx in the integral: $$\int \frac{5 \tan(\theta)}{(\frac{5}{3} \sec(\theta))^3} \frac{5}{3} \sec(\theta) \tan(\theta) d\theta = \int \frac{25}{27}\cot^2(\theta) d \theta$$ To integrate the cotangent squared function, we know that: $$\int \cot^2(\theta) d\theta = -\cot(\theta) - \theta + C$$ So, the integral becomes: $$\int \frac{25}{27} \cot^2(\theta) d\theta = -\frac{25}{27} (\cot(\theta) + \theta) + C$$

Convert back to x

To convert the result back to x, we recall that \(x = \frac{5}{3} \sec(\theta)\). Thus, we have: $$\cos(\theta) = \frac{3}{5} \frac{1}{x}$$ And then, $$\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - (\frac{3}{5} \frac{1}{x})^2} = \frac{1}{x} \sqrt{\frac{25x^2 - 9}{25}}$$ Thus, the cotangent function is given by: $$\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{3}{x} \frac{\sqrt{25x^2 - 9}}{5}$$ Now substitute this back into the integral result: $$-\frac{25}{27} (\frac{3}{x} \frac{\sqrt{25x^2 - 9}}{5} + \theta) + C$$

Final answer

The final answer for the given integral is: $$\int \frac{\sqrt{9x^2 - 25}}{x^3} dx = -\frac{25}{27} (\frac{3}{x} \frac{\sqrt{25x^2 - 9}}{5} + \theta) + C$$

Key concepts

Integrals

Integrals are fundamental concepts in calculus, representing the accumulation of quantities and the area under a curve. They are the reverse process of differentiation, and they help us understand how a function