Problem 52 - Calculus Single Variable

Question

Use trigonometric identities to evaluate \(\int_{0}^{\pi / 2} \cos ^{2}(x / 2) d x\) and \(\int_{0}^{\pi} \sin ^{2}(x) d x\).

Step-by-Step Solution

Answer

\( \int_{0}^{\pi / 2} \cos^{2}(x/2) \ dx = \frac{\pi}{4} + \frac{1}{2} \) and \( \int_{0}^{\pi} \sin^{2}(x) \ dx = \frac{\pi}{2} \).

Step 1: Use a Trigonometric Identity for Simplification

For the integral \( \int_{0}^{\pi / 2} \cos^{2}(x/2) \ dx \), use the identity \( \cos^{2} \theta = \frac{1 + \cos 2\theta}{2} \) by substituting \( \theta = x/2 \). The integral becomes \( \int_{0}^{\pi / 2} \frac{1 + \cos(x)}{2} \ dx \), which simplifies to \( \frac{1}{2} \int_{0}^{\pi / 2} (1 + \cos(x)) \ dx \).

Step 2: Integrate the Simplified Expression

Calculate the integral, \( \frac{1}{2} \int_{0}^{\pi / 2} (1 + \cos(x)) \ dx \). This splits into \( \frac{1}{2} \left( \int_{0}^{\pi / 2} 1 \ dx + \int_{0}^{\pi / 2} \cos(x) \ dx \right) \). First, \( \int_{0}^{\pi / 2} 1 \ dx = \left[x\right]_{0}^{\pi / 2} = \frac{\pi}{2} \). Second, \( \int_{0}^{\pi / 2} \cos(x) \ dx = \left[\sin(x)\right]_{0}^{\pi / 2} = 1 \). Thus, the integral evaluates to \( \frac{1}{2} \left( \frac{\pi}{2} + 1 \right) = \frac{\pi}{4} + \frac{1}{2} \).

Step 3: Apply the Identity to Second Integral

For \( \int_{0}^{\pi} \sin^{2}(x) \ dx \), use \( \sin^{2}\theta = \frac{1 - \cos 2\theta}{2} \). Thus, \( \int_{0}^{\pi} \sin^{2}(x) \ dx = \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \ dx \), which simplifies to \( \frac{1}{2} \int_{0}^{\pi} (1 - \cos(2x)) \ dx \).

Step 4: Perform Integration of the Reduced Form

Calculate \( \frac{1}{2} \left( \int_{0}^{\pi} 1 \ dx - \int_{0}^{\pi} \cos(2x) \ dx \right) \). First, \( \int_{0}^{\pi} 1 \ dx = \pi \). For the second term, \( \int_{0}^{\pi} \cos(2x) \ dx = \left[\frac{1}{2}\sin(2x)\right]_{0}^{\pi} = 0 \) since \( \sin(2\pi) = \sin(0) = 0 \). Thus, the integral evaluates to \( \frac{1}{2}(\pi - 0) = \frac{\pi}{2} \).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in simplifying integrals that involve trigonometric functions like sine and cosine. They allow us to rewrite complex trigonometric expressions into more manageable forms. For example, in the original exercises, we use the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \) to transform the integral \( \int_{0}^{\pi / 2} \cos^{2}(x/2) \ dx \) into a simpler form.

This transformation is key because the cosine squared term is tough to integrate directly. By using the identity, we convert it into a combination of simpler terms: \( 1 \) and \( \cos(x) \). These new terms are straightforward to integrate using basic integration methods.

• \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \)
• \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \)

Identities like these exploit the periodic nature of trigonometric functions and are immensely useful in integration and solving equations. Understanding and applying them correctly can transform a challenging calculus problem into a more manageable one.

Integration Techniques

Integration involves finding the antiderivative of a function, and different techniques are often employed to handle various kinds of integrands. In trigonometric integration, one of the main methods is using substitution and simplification via identities.

After applying a trigonometric identity, as seen in the exercise, we simplify the integral of \( \int_{0}^{\pi / 2} \cos^{2}(x/2) \ dx \) into a sum: \( \frac{1}{2} \int_{0}^{\pi / 2} (1 + \cos(x)) \ dx \). Each part of the sum is integrated separately, utilizing standard integral results:

• \( \int \cos(x) \, dx = \sin(x) + C \)
• \( \int 1 \, dx = x + C \)

Substitution techniques help when dealing with composite functions, like \( \cos^2(x/2) \), allowing for the reduction into simpler parts that can be more easily integrated. Integrating separate terms of \( 1 \) and \( \cos(x) \) becomes a straightforward task, highlighting how breaking down a complex integral into familiar components makes the process much clearer.

Definite Integrals

Definite integrals calculate the net area under a curve over a specified interval, providing a numeric value that represents this area. They are foundational in understanding the total accumulation of a quantity, like distance, area, or volume.

In this exercise, we calculate definite integrals over specific intervals: \([0, \pi/2]\) for \( \cos^2(x/2) \) and \([0, \pi]\) for \( \sin^2(x) \). Each integral involves evaluating at both bounds and finding the difference:

• \( \int_a^b f(x) \, dx = F(b) - F(a) \)

For\( \int_{0}^{\pi / 2} \cos^2(x/2) \ dx \), after the integration, substitute the bounds into the antiderivative to get the result. Similarly, for \( \int_{0}^{\pi} \sin^2(x) \ dx \), the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) simplifies the function before integrating over the interval \([0, \pi]\).

The process of performing definite integrals not only involves finding the antiderivative but also accurately evaluating the function at given limits, which yields important numerical insights in applied mathematics and sciences.