Problem 45 - Calculus Single Variable

Question

Each of the integrands involves an expression of the form \(a^{2}-b^{2} x^{2}, a^{2}+b^{2} x^{2},\) or \(b^{2} x^{2}-a^{2} .\) Use an indirect substitution of the form \(x=(a / b) \sin (\theta), x=(a / b)\) \(\tan (\theta),\) or \(x=(a / b) \sec (\theta)\) to calculate the given integral. $$ \int_{1}^{2} \frac{10}{\sqrt{25 x^{2}-16}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to approximately 1.74.

Step 1: Identify the form

The integrand's denominator is \(\sqrt{25x^2 - 16}\), which matches the form \(b^2 x^2 - a^2\) where \(b^2 = 25\) and \(a^2 = 16\).

Step 2: Choose the substitution

For the form \(b^2 x^2 - a^2\), we use the substitution \(x = \frac{a}{b} \sec(\theta)\). Here, \(a = 4\) and \(b = 5\), so we have \(x = \frac{4}{5} \sec(\theta)\).

Step 3: Differentiate the substitution

Differentiate \(x = \frac{4}{5} \sec(\theta)\) with respect to \(\theta\): \(dx = \frac{4}{5} \sec(\theta) \tan(\theta)\ d\theta\).

Step 4: Substitute in the integral

Substitute \(x\) and \(dx\) in the integral: \(\int \frac{10}{\sqrt{25x^2 - 16}}\ dx = \int \frac{10}{\sqrt{25\left(\frac{4}{5}\sec(\theta)\right)^2 - 16}} \cdot \frac{4}{5} \sec(\theta) \tan(\theta) \ d\theta\).

Step 5: Simplify inside the integral

Simplify the inside of the square root: \(25\left(\frac{4}{5} \sec(\theta)\right)^2 - 16 = 16 \sec^2(\theta) - 16 = 16(\sec^2(\theta) - 1) = 16 \tan^2(\theta)\).

Step 6: Resolve the integral

Now the integral becomes: \(\int \frac{10 \cdot \frac{4}{5} \sec(\theta) \tan(\theta)}{4 \tan(\theta)} d\theta\), which simplifies to \(\int \sec(\theta) d\theta\).

Step 7: Integrate and back-substitute

The integral of \(\sec(\theta)\) is \(\ln|\sec(\theta) + \tan(\theta)| + C\). Replace \(\sec(\theta)\) and \(\tan(\theta)\) back with \(x\) terms: \(\sec(\theta) = \frac{5x}{4}\) and \(\tan(\theta) = \sqrt{\frac{25x^2 - 16}{16}}\).

Step 8: Apply the limits

Evaluate from 1 to 2: Calculate \(\ln|\frac{5x}{4} + \sqrt{\frac{25x^2 - 16}{16}}|\) at \(x = 2\) and \(x = 1\), then find their difference to get the definite integral result.

Key Concepts

Understanding Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals and their properties. Essentially, it is concerned with the accumulation of quantities and the calculation of areas under curves. It provides a systematic way to answer the question 'what is the total quantity accumulated by a varying quantity?' which is critical in solving physical and geometrical problems.
Integrals are classified mainly into two types: indefinite and definite integrals. A definite integral has limits of integration and calculates the net area between the curve and the x-axis over a specified interval. Unlike indefinite integrals, definite integrals produce a numeric result rather than a function. This exercise uses definite integrals to calculate the total accumulated area between specified bounds.
Mathematicians often tackle complex integrals like the one in the exercise by using specific techniques like trigonometric substitution, which simplifies the integration process through clever substitution methods.

Exploring Definite Integrals

A definite integral is essentially an extension of calculating the area under a curve between two points, say from point A to point B on the x-axis. This is performed using the Riemann sum approach in its simpler form, but for practical purposes, we often use the power of calculus.
In the given exercise, the integral \[ \int_{1}^{2} \frac{10}{\sqrt{25 x^{2}-16}} \, dx \] is evaluated between the limits 1 and 2. This describes the accumulation of the function's values between these points, providing a single number or value that represents the aggregate measure.
Such calculations can aid in various applications, from determining distances and areas to understanding the accumulation of quantities over an interval in the real world. Evaluating a definite integral involves using antiderivatives (or the Fundamental Theorem of Calculus) and substituting the upper and lower bounds to find the precise 'amount' being measured.

Mastering Integration Techniques

Integration techniques are fundamental tools to solve complex integrals that are not easily tackled using basic formulas. One such technique is **trigonometric substitution**, which is especially useful when the integrand contains quadratic expressions under a square root, as it does in this exercise.
In our problem, the square root of a quadratic form \(25x^2 - 16\) is present, suggesting the use of a substitution. By recognizing the pattern as \(b^2 x^2 - a^2\), where \(b = 5\) and \(a = 4\), we use the substitution \(x = \frac{4}{5} \sec(\theta)\).
Trigonometric substitution involves substituting variables with trigonometric functions to simplify integration. Post substitution and differentiation, the integral simplifies in trigonometric terms, here transforming into \(\int \sec(\theta) d\theta\). This allows for a more direct integration using known integral formulas. After solving the integral, we back-substitute to express the result in the original variable terms before applying the limits.
These techniques significantly mitigate the complexity of integral calculus, empowering mathematicians and engineers to solve intricate integrals in various practical scenarios like physics, engineering, and beyond.