Problem 52 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(39-56\) $$ \int_{\pi / 4}^{\pi / 2} \cot t d t $$

Step-by-Step Solution

Answer

The value of the integral is \(\frac{1}{2}\ln(2)\).

Step 1: Recall definitions of trigonometric functions

The cotangent function, \(\cot t\), can be rewritten in terms of sine and cosine as \(\cot t = \frac{\cos t}{\sin t}\). This identity will help us to simplify the integral into a more recognizable form.

Step 2: Set up the integral with trigonometric identity

Replace \(\cot t\) in the integral with its equivalent expression \(\frac{\cos t}{\sin t}\). This gives us the new integral: \[ \int_{\pi/4}^{\pi/2} \frac{\cos t}{\sin t} \, dt. \]

Step 3: Use substitution method

Let \(u = \sin t\) which implies that \(du = \cos t \, dt\). Adjust the bounds for \(u\): when \(t = \pi/4\), \(u = \sin(\pi/4) = \frac{\sqrt{2}}{2}\) and when \(t = \pi/2\), \(u = \sin(\pi/2) = 1\). Substitute these into the integral: \[ \int_{\sqrt{2}/2}^{1} \frac{1}{u} \, du. \]

Step 4: Evaluate the integral using basic integration rule

The integral of \(\frac{1}{u}\) with respect to \(u\) is \(\ln |u|\). Thus, the integral becomes: \[ [\ln |u|]_{\sqrt{2}/2}^{1}. \]

Step 5: Calculate the definite integral

Substitute the updated bounds into the evaluated integral: \[ \ln |1| - \ln \left|\frac{\sqrt{2}}{2}\right| = 0 - \left(-\ln \left(\frac{2}{\sqrt{2}}\right)\right) = \ln \left(\sqrt{2}\right). \]

Step 6: Simplify the result

Use the property of logarithms \(\ln(a^b) = b\ln(a)\), where \(\ln(\sqrt{2}) = \frac{1}{2}\ln(2)\). This simplifies to \(\frac{1}{2}\ln(2)\).

Key Concepts

Trigonometric identities

Trigonometric identities are essential tools in calculus, especially when dealing with integrals involving trigonometric functions. They allow us to transform complex expressions into simpler forms that are usually easier to integrate. In this exercise, we use the identity

• \( \cot t = \frac{\cos t}{\sin t} \).

This re-expression of the cotangent function in terms of sine and cosine provides a more manageable function for integration.
By expressing \( \cot(t) \) as a ratio of cosine and sine, we make it easier to apply substitution techniques, which simplify the integral process and help in evaluation. Remembering and correctly applying these identities can significantly streamline solving trigonometric integrals.

Substitution method

The substitution method is a powerful technique for solving integrals, particularly when faced with complex trigonometric integrals. The approach involves replacing a complicated part of the integrand with a single variable. In this example, we let

• \(u = \sin t \),
• \(du = \cos t \, dt \).

The substitution simplifies the integral, transforming it from a trigonometric function to a more straightforward form:\[ \int \frac{\cos t}{\sin t} \, dt \rightarrow \int \frac{1}{u} \, du \]Moreover, don't forget to adjust the limits of integration accordingly. When substituting, compute the values of \(u\) at the original bounds of \(t\). This step is crucial to correctly evaluating the definite integral. Mastering substitution will make evaluating integrals involving trigonometric functions much more manageable.

Definite integrals

Definite integrals calculate the net area between a function and the x-axis within a given interval. They differ from indefinite integrals because they have upper and lower limits. For our problem, the integral is evaluated from

• \(t = \frac{\pi}{4} \) to \(t = \frac{\pi}{2} \).

After substitution, the limits change accordingly:\[\int_{\pi/4}^{\pi/2} \cot t \, dt \rightarrow \int_{\sqrt{2}/2}^{1} \frac{1}{u} \, du\]This process turns the problem into evaluating the definite integral of a simpler function, \(\frac{1}{u}\), within the new bounds.
Always ensure that you update and use these new limits correctly when calculating a definite integral. They are key to obtaining the final numerical result.

Logarithmic integration

Logarithmic integration is fundamental in calculus when dealing with integrals of the form \(\frac{1}{u}\). The result of such an integral is the natural logarithm of the absolute value of \(u\). In this exercise, integrating \(\frac{1}{u}\) gives us

• \( \ln |u| \).

Applying this to our interval, we have:\[[\ln |u|]_{\sqrt{2}/2}^{1} = \ln |1| - \ln \left|\frac{\sqrt{2}}{2}\right|\]Remember that \(\ln(1) = 0\), simplifying the expression to \[- \ln \left(\frac{2}{\sqrt{2}}\right) = \ln(\sqrt{2})\]Using properties of logarithms, this can simplify further to \(\frac{1}{2}\ln(2)\). Mastering logarithmic integration and properties of logarithms widens the scope of integrals you can easily solve.