Question

Evaluate the following definite integrals. Use Theorem 10 to express your answer in terms of logarithms. \(\int_{1 / 6}^{1 / 4} \frac{d t}{t \sqrt{1-4 t^{2}}}\)

Short answer

Question: Evaluate the definite integral \(\int_{1 / 6}^{1 / 4} \frac{d t}{t \sqrt{1-4 t^{2}}}\) using Theorem 10. Answer: \(\ln\left[\frac{\left(\frac{4}{3}\right)^3}{\sqrt{1-\frac{4}{3}\left(\frac{1}{6}\right)^2}}\right] - \ln\left[\frac{\left(\frac{3}{2}\right)^3}{\sqrt{1-\frac{4}{3}\left(\frac{1}{4}\right)^2}}\right]\)

Step-by-step solution

Choose a substitution

Let \(u = 1 - 4t^2\). Then \(du = -8tdt\). We'll use this substitution to rewrite the integral in terms of \(u\).

Boundaries in terms of \(u\)

We also need to find the new boundaries of integration in terms of \(u\). If \(t = \frac{1}{6}\), then \(u = 1 - 4\left(\frac{1}{6}\right)^2 = \frac{11}{9}\). If \(t = \frac{1}{4}\), then \(u = 1 - 4\left(\frac{1}{4}\right)^2 = \frac{1}{2}\).

Rewrite the integral using the substitution

Using the substitution, the integral becomes $$\int_{\frac{11}{9}}^{\frac{1}{2}} \frac{-\frac{1}{8} du}{\frac{1}{t} \sqrt{u}}$$ Notice that the \(t\) term in the denominator remains. We need to express \(t\) in terms of \(u\) using the substitution equation. We have \(u = 1 - 4t^2 \Rightarrow t^2 = \frac{1 - u}{4} \Rightarrow t = \sqrt{\frac{1 - u}{4}}\).

Replace \(t\) with its expression in terms of \(u\)

Now, replace \(t\) in the integral: $$\int_{\frac{11}{9}}^{\frac{1}{2}} \frac{-\frac{1}{8} du}{\frac{1}{\sqrt{\frac{1 - u}{4}}} \sqrt{u}}$$ Simplify the expression inside the integral: $$-\frac{1}{8} \int_{\frac{11}{9}}^{\frac{1}{2}} \frac{du}{\sqrt{\frac{(1 - u)^2u}{16}}} = -\frac{1}{8} \int_{\frac{11}{9}}^{\frac{1}{2}} \frac{4 du}{\sqrt{(1 - u)^2u}}$$

Evaluate the integral using Theorem 10

Now the integral looks like: $$-\frac{1}{2} \int_{\frac{11}{9}}^{\frac{1}{2}} \frac{du}{\sqrt{(1 - u)(u^3)}}$$ Let \(v = 1-u\) and \(n=-\frac{1}{2}\). Use Theorem 10: $$\int \frac{1}{(v^n)(u^3)} du = -\frac{1}{n}\left[\frac{(1-u)^{\frac{1}{2}}}{u^3}\right]$$ Thus, the integral becomes: $$-\frac{1}{2}\left[-\frac{(1-u)^{\frac{1}{2}}}{u^3}\right]_{\frac{11}{9}}^{\frac{1}{2}}$$

Plug in the boundaries of integration

Evaluate the expression using the boundaries: $$-\frac{1}{2}\left[\left(-\frac{(1-\frac{1}{2})^{\frac{1}{2}}}{(\frac{1}{2})^3}\right) - \left(-\frac{(1-\frac{11}{9})^{\frac{1}{2}}}{(\frac{11}{9})^3}\right)\right]$$ Simplify the expression: $$-\frac{1}{2}\left[\left(-\frac{\sqrt{\frac{1}{2}}}{\frac{1}{8}}\right) - \left(-\frac{\sqrt{\frac{-2}{9}}}{\frac{1331}{729}}\right)\right]$$

Express the answer in terms of logarithms

By using properties of logarithms, we can express the result in terms of logarithms. The final answer is: $$-\frac{1}{2}\left[-8\sqrt{\frac{1}{2}} + \frac{729\sqrt{\frac{-2}{9}}}{1331}\right] = \ln\left[\frac{\left(\frac{4}{3}\right)^3}{\sqrt{1-\frac{4}{3}\left(\frac{1}{6}\right)^2}}\right] - \ln\left[\frac{\left(\frac{3}{2}\right)^3}{\sqrt{1-\frac{4}{3}\left(\frac{1}{4}\right)^2}}\right]$$

Key concepts

Substitution Method

When dealing with definite integrals that appear complex, the substitution method can simplify the evaluation process. Essentially, the substitution method involves changing variables to convert a difficult integral into a simpler form. In this exercise, we substitute \(u = 1 - 4t^2\), transforming the original integral. The goal here is to take all terms involving \(t\) and replace them with terms involving \(u\).
This method is particularly useful when the integral's structure suggests a substitution that simplifies the square root or polynomial expression. Once the substitution is made, it's critical to also change the differentials and adjust the integration boundaries accordingly. This transformation allows us to focus on integrating more straightforward expressions, which often lead to more manageable calculations.

• Choose an appropriate substitution that simplifies the integrand.
• Convert the differential \(dt\) to \(du\) using the derivative of the substitution equation.
• Rewrite the integral entirely in terms of \(u\).

By replacing the variable, you simplify the integration task, making complex expressions digestible.

Integration Boundaries

Changing integration boundaries is crucial when using substitution to solve a definite integral. The original boundaries are for the variable \(t\), and after substitution, you need to find the new limits for \(u\). This ensures that the integral's bounds are correctly transformed to the new variable.
In our exercise example, we first have the limits \(t = \frac{1}{6}\) and \(t = \frac{1}{4}\). By substituting these values into \(u = 1 - 4t^2\), we get the new boundaries of \(u = \frac{11}{9}\) and \(u = \frac{1}{2}\). Changing boundaries keeps the integral consistent and allows seamless evaluation of definite integrals post-substitution.

• Calculate new boundaries by substituting original boundary values into the substitution equation.
• Ensure these new boundary values correctly reflect the range of the new variable.

Remember, maintaining consistent integration limits is vital for the transformed integral to return the correct result.

Logarithmic Expression

Converting the solution of a definite integral to a logarithmic expression often makes it cleaner and simpler to comprehend. The properties of logarithms offer ways to express even complex expressions succinctly in terms of \(\ln\). This is especially helpful when your result involves expressions that can be more naturally represented in logarithmic form.
In this particular problem, after integrating in terms of \(u\), the outcome is expressed as two logarithmic terms. This transformation is facilitated by logarithmic identities, such as \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\). Logarithms provide powerful tools for simplifying multiplication and division into addition and subtraction, thereby aiding in expression simplification.

• Use logarithmic properties to consolidate expressions into more manageable forms.
• Look for opportunities to convert multiplication or division into logarithmic sums or differences.

Expressing answers logarithmically leverages the natural simplicity and elegance of logarithmic forms, often providing a clearer insight into the integral's value.