Problem 105 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{\pi / 4} \sin (2 x) d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Integral

We are given the definite integral \( \int_{0}^{\pi / 4} \sin(2x) \, dx \). The task is to evaluate this integral over the interval from 0 to \( \pi/4 \).

Step 2: Set Up the Substitution

To solve the integral, we'll use a substitution to simplify the integration. Let \( u = 2x \). Consequently, \( du = 2 \, dx \), which implies that \( dx = \frac{du}{2} \).

Step 3: Change Limits of Integration

Change the limits of integration according to the substitution \( u = 2x \):- When \( x = 0 \), \( u = 2 \times 0 = 0 \).- When \( x = \pi/4 \), \( u = 2 \times \pi/4 = \pi/2 \).Thus, the new limits for \( u \) are from 0 to \( \pi/2 \).

Step 4: Substitute and Integrate

Substitute \( u \) and \( dx \) into the integral:\[\int_{0}^{\pi/2} \sin(u) \cdot \frac{du}{2} = \frac{1}{2} \int_{0}^{\pi/2} \sin(u) \, du.\]Now, integrate \( \sin(u) \, du \), which results in \( -\cos(u) \). Thus:\[\frac{1}{2} \left[-\cos(u) \right]_{0}^{\pi/2}.\]

Step 5: Evaluate the Antiderivative

Evaluate \( -\cos(u) \) from 0 to \( \pi/2 \):\[\frac{1}{2} \left[ -\cos(\pi/2) + \cos(0) \right].\]Since \( \cos(\pi/2) = 0 \) and \( \cos(0) = 1 \), this becomes:\[\frac{1}{2} \left[ 0 - 1 \right] = \frac{1}{2} \times (-1).\]

Step 6: Simplify the Result

Simplify the expression to find the value of the integral:\[\frac{-1}{2} = -\frac{1}{2}.\]

Key Concepts

Trigonometric Functions

Trigonometric functions, like sine and cosine, appear frequently in calculus due to their periodic nature and diverse applications. In our exercise, we evaluate the integral of the sine function, specifically \( \sin(2x) \). The sine function is essential in modeling waves, oscillations, and circular motion. Each trigonometric function has specific properties; for sine, it is periodic and has a range from \([-1, 1]\).
When integrating different trigonometric functions, we often utilize their derivatives or antiderivatives. For example:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The antiderivative of \( \sin(x) \) is \(-\cos(x) + C\), where \( C \) is the constant of integration.

Knowing these properties aids in efficiently solving integrals involving trigonometric functions.

Integration Techniques

Integration techniques are essential tools for evaluating integrals, especially those that are not straightforward. In this exercise, we learn about regularly used methods like substitution. This technique transforms a complex integral into a simpler form that is easier to handle.
Other common techniques include:

• Integration by parts: Based on the product rule and ideal for integrating the product of functions.
• Partial fraction decomposition: Useful for rational functions to break them into simpler fractions.
• Trigonometric identities: Utilize identities to simplify trigonometric integrals before solving.

Mastering these methods ensures that you can tackle a wide range of integral problems, whether they involve trigonometric, polynomial, or exponential functions.

Substitution Method

The substitution method acts as a bridge to simplify complex integrals by transforming them into a familiar form. In our example, we substitute \( u = 2x \), simplifying integration by removing the coefficient in the argument of the sine function. This substitution impacts both the variable inside the trigonometric function and the differential \( dx \).
The steps are simple:

• Identify a part of the integral that can be substituted to simplify the process.
• Transform the integral and change the limits of integration according to the new variable, here from \( x \) to \( u \).
• Solve the integral in its simpler form.

The substitution method is particularly useful when dealing with composite functions or when derivatives match part of the integrand. By mastering this technique, solving integrals becomes a clearer, more structured process.