Question

\((\pi / 2) \int_{0} \log [\tan (x / 2)+\cot (x / 2)] \mathrm{d} x\) is equal to \(\ldots \ldots\) (a) \((\pi / 2) \log 2\) (b) \(-(\pi / 2) \log 2\) (c) \(\pi \log 2\) (d) \(-\pi \log 2\)

Short answer

The given integral is equal to \(-\pi \log{2}\), so the correct answer is option (d).

Step-by-step solution

Analyze the given integral

The given integral is: \[ (\pi / 2) \int_{0} \log [\tan (x / 2)+\cot (x / 2)] \, \mathrm{d} x \] First, we need to simplify the argument inside the logarithm.

Simplify the argument inside the logarithm

To simplify the argument, we need to recall that: \[ \cot{u} = \frac{1}{\tan{u}} \] Now,let's set \(u=x/2\): \[ \begin{aligned} \tan(u) + \cot(u) & = \tan(u) + \frac{1}{\tan(u)} \\ & = \frac{\tan^{2}(u) + 1}{\tan(u)} \\ \end{aligned} \] Recall that \(\tan^{2}(u) + 1 = \sec^{2}(u)\). So, \[ \frac{\tan^{2}(u) + 1}{\tan(u)} = \frac{\sec^{2}(u)}{\tan(u)} \] Now, let's put \(u\) back in terms of x: \[ \frac{\sec^{2}(x/2)}{\tan(x/2)} = \frac{1}{\sin(x/2)\cos(x/2)} \] Replace the argument inside the logarithm with the simplified expression and write the integral as: \[ (\pi / 2) \int_{0} \log \frac{1}{\sin(x/2)\cos(x/2)} \, \mathrm{d} x \]

Use the logarithm property to simplify the integral

Recall that \(\log \frac{1}{a} = -\log a\). So we have, \[ (\pi / 2) \int_{0}(-\log[\sin(x/2)\cos(x/2)]) \, \mathrm{d} x \] Now, we can move the negative sign outside the integral and use the logarithm properties to simplify the expression further, \[ -(\pi / 2) \int_{0} \log[\sin(x/2)\cos(x/2)] \, \mathrm{d} x = -(\pi / 2) \int_{0}(\log[\sin(x/2)] + \log[\cos(x/2)]) \, \mathrm{d} x \]

Split the integral

Split the integral into two separate integrals, \[ -(\pi / 2) \int_{0} \log[\sin(x/2)] \, \mathrm{d} x - (\pi / 2) \int_{0} \log[\cos(x/2)] \, \mathrm{d} x \] Notice the symmetry between the two integrals. We can use the substitution \(x = \pi - t\) and the property \(\cos(\frac{\pi}{2} - x) = \sin(x)\) to show that the two integrals are equal.

Show that the two integrals are equal

Perform the substitution \(x = \pi - t\), \[ -(\pi / 2)\int_{0}^{\pi} \log[\cos(t/2)] \, \mathrm{d} t = -(\pi / 2) \int_{0}^{\pi} \log[\sin(\pi - t/2)] \, \mathrm{d} t \] As per the property \(\cos(\frac{\pi}{2} - x) = \sin(x)\), we have \[ -(\pi / 2) \int_{0}^{\pi} \log[\sin(\pi - t/2)] \, \mathrm{d} t = -(\pi / 2) \int_{0}^{\pi} \log[\sin(t/2)] \, \mathrm{d} t \] So, adding these equal integrals together, we get \[ -2(\pi / 2) \int_{0}^{\pi} \log[\sin(t/2)] \, \mathrm{d} t \] Rewrite the expression, \[ -\pi \int_{0}^{\pi} \log[\sin(t/2)] \, \mathrm{d} t = -\pi \int_{0}^{\pi} \log[\sin(t/2)] \, \mathrm{d} t \]

Evaluate the integral

After the previous steps, we can rewrite the given integral as the following integral: \[ -\pi \int_{0}^{\pi} \log[\sin(t/2)] \, \mathrm{d} t \] This is a well-known integral related to \(\log{2}\), and the result is: \[ -\pi \log{2} \]

Identify the correct answer

Comparing our result with the given options, we can see that the integral is equal to: \[ -\pi \log{2} \implies \mathrm{(d)} \] Therefore, the correct answer is option (d).

Key concepts

Integration Techniques

When tackling an integral, familiarity with a range of integration techniques is vital for simplifying complex expressions and finding solutions. In this exercise, the integral involves a logarithmic function within a trigonometric context, which seems complex at first glance. To manage this, the problem is broken down into simpler steps, allowing the use of algebraic identities and properties of logarithms to transform the integral into a more manageable form. For example, by applying the identity \( \tan^2(u) + 1 = \text{sec}^2(u) \) and the logarithmic property \( \text{log} \frac{1}{a} = -\text{log} a \) we can vastly simplify the problem. It's also helpful to split the integral into two parts if symmetry is observed, as it can lead to further simplifications. By mastering various techniques, such as trigonometric identities, properties of logarithms, and strategic substitutions, students can solve integrals that initially appear daunting.

Logarithmic Functions

Understanding Logarithms

Logarithmic functions are the inverse operations of exponential functions and play a critical role in many areas of mathematics, including the process of integration. Their properties are often utilized to break down complex expressions, which is demonstrated in this exercise. The property used here is \( \text{log} \frac{1}{a} = -\text{log} a \) which when applied, changes the integrand from a quotient inside a logarithm to a simpler, negative logarithmic expression. This not only simplified the integral but also paved the way to further break it down into two integrals that exploit the symmetry of \text{sin} and \text{cos} functions, a common technique when dealing with trigonometric integrals with logarithms. Familiarity with such log properties is essential for students to simplify and evaluate integrals effectively.

Trigonometric Integrals

Trigonometric integrals involve integrating functions composed of trigonometric functions, which often require a creative application of trigonometric identities for their solution. The exercise provided presents a case where the integral of a logarithmic function of trigonometric expressions is evaluated. One critical aspect of dealing with trigonometric integrals is the recognition of patterns and symmetry, which allows for substitutions that can simplify the integral. This problem uses the symmetry of sine and cosine functions, specifically under the transformation \( x = \text{pi} - t \) to show that two seemingly different integrals are, in fact, equal. This insight is crucial as it reduces the complexity of the problem, ultimately allowing for a straightforward integration. Knowledge of such symmetry in trigonometric functions and strategic substitutions is a powerful tool when solving trigonometric integrals.