Problem 113 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{\pi / 2} \cos 2 x d x $$

Step-by-Step Solution

Answer

The definite integral evaluates to 0.

Step 1: Identifying the Integral

We need to evaluate the definite integral from \( 0 \) to \( \frac{\pi}{2} \) of the function \( \cos 2x \). The integral is given by \( \int_{0}^{\pi / 2} \cos 2x \, dx \).

Step 2: Use a Substitution Method

Let's use substitution to simplify the integral. Let \( u = 2x \). Then, the derivative \( du = 2 \, dx \), which implies \( dx = \frac{1}{2} \, du \). The limits of integration change accordingly. For \( x = 0 \), \( u = 2 \times 0 = 0 \). For \( x = \frac{\pi}{2} \), \( u = 2 \times \frac{\pi}{2} = \pi \). Thus, the integral becomes \( \int_{0}^{\pi} \cos u \, \frac{1}{2} \, du \).

Step 3: Simplify the Integral

Factor out the constant \( \frac{1}{2} \) from the integral. The expression becomes \( \frac{1}{2} \int_{0}^{\pi} \cos u \, du \).

Step 4: Evaluate the Integral

The integral of \( \cos u \) is \( \sin u \). Thus, \( \frac{1}{2} \int_{0}^{\pi} \cos u \, du = \frac{1}{2} [\sin u]_{0}^{\pi} \).

Step 5: Substitute the Limits of Integration

Substitute the limits into the result from Step 4: \[ \frac{1}{2} (\sin \pi - \sin 0) = \frac{1}{2} (0 - 0) = 0 \]

Step 6: State the Final Answer

The definite integral evaluates to \( 0 \). Thus, \( \int_{0}^{\pi / 2} \cos 2x \, dx = 0 \).

Key Concepts

Understanding the Substitution Method

The substitution method is a fundamental technique in calculus for evaluating integrals. It simplifies complex integrals by transforming the variable and integrand into a simpler form. Consider our function \( \cos 2x \), which we aim to integrate. We select a substitution, such as \( u = 2x \), to simplify this process.

• First, we find the derivative of \( u = 2x \) to get \( du = 2 \, dx \).
• Rearrange to solve for \( dx \), yielding \( dx = \frac{1}{2} \, du \).
• The integration variable shifts from \( x \) to \( u \), transforming our original integral to involve \( u \) instead.

This substitution rephrases the integral to \( \int_{0}^{\pi} \cos u \, \frac{1}{2} \, du \). This simplifies the integrand and allows simpler evaluation, emphasizing the power of the substitution method in tackling trigonometric functions.

Applying Integration Limits

Integration limits are critical in evaluating definite integrals as they define the region of accumulation. With the substitution \( u = 2x \), the original limits need revision:
- For \( x = 0 \), substituting in gives \( u = 2 \times 0 = 0 \).- For \( x = \frac{\pi}{2} \), substituting in results in \( u = 2 \times \frac{\pi}{2} = \pi \).These recalibrated limits transform our integration bounds from \( [0, \frac{\pi}{2}] \) with respect to \( x \) to \( [0, \pi] \) with respect to \( u \).
Changing limits ensures the integrity and correctness of the solution despite modifying the integration variable. This ensures continuity and correctness in both definite and indefinite integrals.

Evaluating Trigonometric Integrals

Trigonometric integrals often involve functions like sine and cosine, utilized frequently in physics and engineering. Evaluating them requires knowledge of their antiderivatives:

• The integral of \( \cos u \) with respect to \( u \) is \( \sin u \).
• Thus, \( \int \cos u \, du = \sin u + C \), where \( C \) is a constant for indefinite integrals.

For definite integrals, apply the Newton-Leibniz theorem, which states:

1. Find the antiderivative, \( \sin u \), of the function.
2. Evaluate \( [\sin u]_{0}^{\pi} = \sin \pi - \sin 0 \).
3. Since both \( \sin \pi \) and \( \sin 0 \) are 0, the evaluated integral is \( 0 \).

The trigonometric function's periodic nature, with known sine and cosine values at critical angles, simplifies evaluation, providing an elegant solution.