Given, $i=\sqrt{-1}$, the value of the definite integral, $I=\underset{0}{\overset{\pi /2}{\int }}\frac{\cos x+i\sin x}{\cos x-i\sin x}dx$ is:

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:

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{\cos x+i\sin x}{\cos x-i\sin x}dx$, 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:

$I=\underset{0}{\overset{\pi /2}{\int }}{e}^{2ix}dx$

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:

$\int e^{2ix}dx=\frac{1}{2i}{e}^{2ix}$

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, $I=\underset{0}{\overset{\pi /2}{\int }}\frac{\cos x+i\sin x}{\cos x-i\sin x}dx$, is $i$.