Definite Integral Solution with Complex Numbers

This problem asks us to find the value of a definite integral involving complex numbers. We are given the integral:

\(I = \mathop \smallint \limits_0^{\pi /2} \frac{{\cos x + i\sin x}}{{\cos x - i\sin x}}dx\), where \(i = \sqrt { - 1} \).

Euler's Formula: Key Concept for Complex Integrals

The core of solving this definite integral lies in using Euler's formula, which establishes a fundamental relationship between trigonometric functions and the complex exponential function. This formula is crucial for simplifying expressions involving complex numbers in trigonometric form:

• \(\cos \theta + i\sin \theta = e^{i\theta}\)

We will apply this formula to simplify the integrand of the definite integral before proceeding with the integration.

Integral Calculation: Step-by-Step Process

Integrand Simplification using Euler's Formula

First, let's simplify the expression inside the definite integral. The numerator is \(\cos x + i\sin x\). Applying Euler's formula directly, we get:

• Numerator: \(\cos x + i\sin x = e^{ix}\)

For the denominator, \(\cos x - i\sin x\), we can use the properties of trigonometric functions where \(\cos(-x) = \cos x\) and \(\sin(-x) = -\sin x\). So, we can rewrite it as:

• \(\cos x - i\sin x = \cos(-x) + i\sin(-x)\)

Now, applying Euler's formula to this form gives us:

• Denominator: \(\cos(-x) + i\sin(-x) = e^{-ix}\)

Substitute these simplified forms back into the fraction of the integrand:

\(\frac{{\cos x + i\sin x}}{{\cos x - i\sin x}} = \frac{e^{ix}}{e^{-ix}}\)

Using the properties of exponents, specifically \(\frac{a^m}{a^n} = a^{m-n}\), we simplify the expression further:

\(\frac{e^{ix}}{e^{-ix}} = e^{ix - (-ix)} = e^{ix + ix} = e^{2ix}\)

Thus, the original definite integral transforms into a simpler form:

\(I = \mathop \smallint \limits_0^{\pi /2} e^{2ix} dx\)

Integration Process for the Complex Exponential

Now we need to perform the integration of \(e^{2ix}\) with respect to \(x\). The general rule for integrating \(e^{ax}\) is \(\frac{1}{a}e^{ax}\). In our case, \(a = 2i\).

So, the indefinite integral is:

\(\smallint e^{2ix} dx = \frac{1}{2i} e^{2ix}\)

Since we are dealing with a definite integral, we do not need to include the constant of integration.

Limits of Integration Application

Next, we apply the upper and lower limits of integration, \(\pi/2\) and \(0\) respectively, to the antiderivative:

\(I = \left[ \frac{1}{2i} e^{2ix} \right]_0^{\pi /2}\)

\(I = \frac{1}{2i} \left( e^{2i(\pi/2)} - e^{2i(0)} \right)\)

\(I = \frac{1}{2i} \left( e^{i\pi} - e^0 \right)\)

Exponential Terms Evaluation for the Definite Integral

To finalize the calculation, we need to evaluate the exponential terms \(e^{i\pi}\) and \(e^0\). We will again use Euler's formula for \(e^{i\pi}\).

Evaluating \(e^{i\pi}\):

• Using Euler's formula, \(e^{i\pi} = \cos \pi + i\sin \pi\).
• We know that \(\cos \pi = -1\) and \(\sin \pi = 0\).
• Therefore, \(e^{i\pi} = -1 + i(0) = -1\).

Evaluating \(e^0\):

• Any non-zero number raised to the power of zero is 1.
• So, \(e^0 = 1\).

Integral Value Calculation

Now, substitute these evaluated values back into our expression for \(I\):

\(I = \frac{1}{2i} (-1 - 1)\)

\(I = \frac{1}{2i} (-2)\)

\(I = \frac{-2}{2i}\)

\(I = \frac{-1}{i}\)

To present the final answer in a standard form (rationalizing the denominator), we multiply the numerator and denominator by \(i\):

\(I = \frac{-1}{i} \times \frac{i}{i}\)

\(I = \frac{-i}{i^2}\)

Since \(i^2 = -1\) (by definition, as \(i = \sqrt{-1}\)):

\(I = \frac{-i}{-1}\)

\(I = i\)

Final Integral Value

The value of the definite integral, \(I = \mathop \smallint \limits_0^{\pi /2} \frac{{\cos x + i\sin x}}{{\cos x - i\sin x}}dx\), is \(i\).