Question

Perform the indicated integrations. $$ \int_{0}^{\pi / 2} \cos ^{5} \theta d \theta $$

Short answer

The integral evaluates to \( \frac{8}{15} \).

Step-by-step solution

Understand the Problem

We need to evaluate the definite integral \( \int_{0}^{\pi/2} \cos^5 \theta \, d\theta \). This means computing the integral of \( \cos^5 \theta \) from \( \theta = 0 \) to \( \theta = \pi/2 \).

Use Trigonometric Identity

Recall that \( \cos^2 \theta = 1 - \sin^2 \theta \). We can write \( \cos^5 \theta = \cos^4 \theta \cdot \cos \theta = (\cos^2 \theta)^2 \cdot \cos \theta = (1 - \sin^2 \theta)^2 \cdot \cos \theta \).

Make a Substitution

Let \( u = \sin \theta \), then \( du = \cos \theta \, d\theta \). The limits of integration change from \( \theta = 0 \) to \( u = \sin 0 = 0 \), and from \( \theta = \pi/2 \) to \( u = \sin(\pi/2) = 1 \). The integral becomes \( \int_{0}^{1} (1 - u^2)^2 \, du \).

Expand and Integrate

Expand the integrand: \( (1 - u^2)^2 = 1 - 2u^2 + u^4 \). Integrate term-by-term: \[ \int_{0}^{1} (1 - 2u^2 + u^4) \, du = \left[ u - \frac{2}{3}u^3 + \frac{1}{5}u^5 \right]_{0}^{1} \].

Evaluate the Definite Integral

Apply the limits to find the value of the definite integral: \( \left[ 1 - \frac{2}{3}\times1^3 + \frac{1}{5}\times1^5 \right] - \left[ 0 - \frac{2}{3}\times0^3 + \frac{1}{5}\times0^5 \right] = 1 - \frac{2}{3} + \frac{1}{5} \).

Simplify the Result

Calculate the expression: \( 1 - \frac{2}{3} + \frac{1}{5} = \frac{15}{15} - \frac{10}{15} + \frac{3}{15} = \frac{8}{15} \). Thus, the integral evaluates to \( \frac{8}{15} \).

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in integration to simplify expressions involving trigonometric functions. In our exercise, we have the integral \( \int_{0}^{\pi/2} \cos^5 \theta \, d\theta \). At first glance, integrating such a complex trigonometric form directly can be challenging.

Here, we use a substitution strategy by recognizing that \( \cos^2 \theta \) can be expressed using \( \sin^2 \theta \) easily through the identity \( \cos^2 \theta = 1 - \sin^2 \theta \). Therefore, substituting \( u = \sin \theta \) we utilized the derivative \( du = \cos \theta \, d\theta \), greatly simplifying our integration process.

This substitution transforms a potentially complex integral into a simpler polynomial form, which becomes manageable through straightforward integration techniques.

Integral Limits

Integral limits are crucial in definitive integration as they define the interval over which the function is to be integrated. Our problem involves the integral \( \int_{0}^{\pi/2} \cos^5 \theta \, d\theta \), with limits of integration from 0 to \( \pi/2 \). This interval defines the range in which the integral must be evaluated.

When we perform trigonometric substitution by setting \( u = \sin \theta \), the boundaries change with respect to \( u \). At \( \theta = 0 \), \( u = \sin 0 = 0 \), and at \( \theta = \pi/2 \), \( u = \sin(\pi/2) = 1 \). These new limits transform the original integral into \( \int_{0}^{1} (1 - u^2)^2 \, du \).

Adapting the limits accurately ensures the integration covers the same interval, just in terms of the new variable \( u \). This is a fundamental aspect of maintaining equivalent integration bounds through substitutions.

Trigonometric Identities

Trigonometric identities play a critical role in simplifying integration problems involving trigonometric functions. Such identities enable us to transform complex expressions into more manageable forms. In our exercise, the identity \( \cos^2 \theta = 1 - \sin^2 \theta \) is especially useful. It helps us rewrite \( \cos^5 \theta \) as \( (1 - \sin^2 \theta)^2 \cdot \cos \theta \).

By rewiring trigonometric powers in terms of sine and cosine, these identities facilitate subsequent manipulative steps. In many cases, they help convert intricate trigonometric integrals into polynomials, which are much easier to work with. Recognizing and applying these identities effectively is key to simplifying and solving integrals involving trigonometric functions.

Integration Techniques

Numerous integration techniques can be applied to solve definite integrals, and our example showcases several of them effectively. After using trigonometric substitution and changing the integral's limits, we transform the integral into \( \int_{0}^{1} (1 - u^2)^2 \, du \).

Now, expanding \( (1 - u^2)^2 \) into \( 1 - 2u^2 + u^4 \), we can easily integrate each term separately. This process, known as term-by-term integration, involves applying standard integration formulas: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C\).

After finding the antiderivatives, we apply the limits of integration to compute the exact area under the curve. Lastly, it's a good practice to simplify the result into a neat fraction, as evidenced by our solution: \( \frac{8}{15} \). Mastering these fundamental techniques helps tremendously in solving other trigonometric integrals.