Question

Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms. $$\int_{1 / 6}^{1 / 4} \frac{d t}{t \sqrt{1-4 t^{2}}}$$

Short answer

Based on the given step by step solution, provide a short answer to the problem: To find the definite integral $$\int_{1 / 6}^{1 / 4} \frac{dt}{t \sqrt{1 - 4t^{2}}}$$, we used substitution and then applied Theorem 7.7 to express the result in terms of logarithms. The final simplified result is: $$\int_{1 / 6}^{1 / 4} \frac{dt}{t \sqrt{1 - 4t^{2}}} = 2\left[\ln\frac{(3 + \sqrt{8})}{(2 + \sqrt{3})}\right]$$

Step-by-step solution

Substitution

First, let's perform a substitution. Let $$x = 2t$$, then, $$dx = 2dt$$ and the bounds of the integral will change as well. When $$t = 1/6$$, we have $$x = 1/3$$, and when $$t = 1/4$$, we have $$x = 1/2$$. The integral now becomes: $$ \int_{1 / 3}^{1 / 2} \frac{d x}{(x / 2) \sqrt{1 - x^{2}}} $$ Now divide by the constant, which gives us $$ \int_{1 / 3}^{1/ 2} \frac{2 d x}{x \sqrt{1 - x^{2}}} $$

Use Theorem 7.7

We recognize that the integral is of the form $$\int_{a}^{b} \frac{d x}{x \sqrt{1 - x^{2}}}$$, and according to Theorem 7.7, we can rewrite this integral in terms of logarithms. The formula that we can use for doing this is: $$\int\frac{dx}{x\sqrt{1-x^2}}=\ln⁡{\frac{x}{1+\sqrt{1-x^2}}}+C$$ Applying this formula to our definite integral: $$ \int_{1 / 3}^{1 / 2} \frac{2 d x}{x \sqrt{1 - x^{2}}} = 2\left[\ln\left( \frac{1/2}{1 + \sqrt{1 - (1/2)^2}}\right) - \ln\left( \frac{1/3}{1 + \sqrt{1 - (1/3)^2}}\right)\right] $$

Simplifying

Now let's simplify the expression: $$ 2\left[\ln\left( \frac{1/2}{1 + \sqrt{1 - (1/2)^2}}\right) - \ln\left( \frac{1/3}{1 + \sqrt{1 - (1/3)^2}}\right)\right] = 2\left[\ln\left(\frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}}\right) - \ln\left(\frac{\frac{1}{3}}{1 + \frac{\sqrt{8}}{3}}\right)\right] $$ So, we have $$ 2\left[\ln\left(\frac{1}{2 + \sqrt{3}}\right) - \ln\left(\frac{1}{3 + \sqrt{8}}\right)\right] $$ Which we rewrite as $$2\left[\ln{\frac{(3 + \sqrt{8})}{(2 + \sqrt{3})}}\right]$$ So, $$ \int_{1 / 6}^{1 / 4} \frac{dt}{t \sqrt{1 - 4t^{2}}} = 2\left[\ln\frac{(3 + \sqrt{8})}{(2 + \sqrt{3})}\right] $$

Key concepts

Integration by Substitution

Integration by substitution is a fundamental technique in calculus, often used to simplify complex integrals. The core idea is to transform the original integral into a more manageable form by changing variables. This method is analogous to finding a substitution in algebra that makes an equation easier to solve.

Here's a general approach to integration by substitution:

• Choose a substitution that simplifies the integrand.
• Replace the variable of integration and differentials in the original integral with the new variables.
• Adjust the limits of integration if dealing with a definite integral.
• Integrate the simpler expression.
• Resubstitute the original variables back into the final expression if necessary.

In the example given, the substitution of \( x = 2t \) simplifies the integral's core expression, which makes it possible to apply further integration techniques, such as logarithmic integration.

Logarithmic Integration

Logarithmic integration is employed when the integrand involves a quotient featuring a polynomial and its derivative, leading to a natural logarithm result. When integrating functions of the form \(\frac{1}{x f'(x)}\), where \(f(x)\) is a function whose derivative \(f'(x)\) is also present in the integrand, the integral often simplifies to the natural logarithm of the absolute value of \(f(x)\).

For this problem, after applying the substitution, the integrand has a structure that fits the pattern for logarithmic integration. Hence, the integral transforms into a logarithmic expression involving the square root of a quadratic, further simplified using logarithm properties. Understanding the direct application of logarithmic identities is crucial to compute these integrals accurately.

Calculus Theorem 7.7

Calculus Theorem 7.7 refers to a specific theorem in calculus that provides a way to express certain types of integrals. While the theorem could vary depending on the textbook, it usually pertains to integrals involving inverse functions or specific substitution patterns leading to logarithmic expressions.

In the context of this exercise, Theorem 7.7 enables the student to express the given integral in terms of logarithms directly without the intermediate integration steps usually required. Awareness of such theorems and the ability to recognize when they apply can greatly expedite solving integrals. This particular theorem offers a shortcut to integrating expressions that involve a variable within a square root, often leading to an inverse trigonometric or logarithmic function as the antiderivative.

Indefinite Integral

An indefinite integral, or antiderivative, represents a general form of all possible antiderivatives of a function. It comes with a '+ C' to indicate that there are an infinite number of functions that can be derived to give the original function, each differing by a constant. Indefinite integrals are used to find the general form of the antiderivative without considering specific limits.

When working with indefinite integrals, the focus is not on calculating the area under a curve, but on finding a function that represents all possible functions whose derivative equals the original integrand. This problem initially requires calculating a definite integral, but understanding indefinite integrals is useful for grasping the broader concept of integration, including integration by substitution and logarithmic integration. Therefore, appreciating the difference between definite and indefinite integrals is essential for a thorough understanding of integral calculus.