Problem 5 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-14\) $$ \int_{0}^{\pi / 2} \sin ^{7} y d y $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{16}{35} \).

Step 1: Identify the Type of Integral

The integral provided is \( \int_{0}^{\pi / 2} \sin ^{7} y \, dy \). It is an integral of a trigonometric function raised to an odd power, which typically suggests the use of a reduction formula or a substitution to simplify it.

Step 2: Use Trigonometric Identity to Simplify

Since the power is odd, we can use the identity \( \sin^2 y = 1 - \cos^2 y \). Rewrite \( \sin^7 y \) as \( \sin^6 y \cdot \sin y = (1 - \cos^2 y)^3 \cdot \sin y \).

Step 3: Apply Substitution

Let \( u = \cos y \), which implies \( du = -\sin y \, dy \). The limits change accordingly as \( y = 0 \) implies \( u = 1 \), and \( y = \pi/2 \) implies \( u = 0 \). The integral becomes \(-\int_{1}^{0} (1 - u^2)^3 \, du \).

Step 4: Adjust Limits and Integrate

Reverse the limits and remove the negative sign: \( \int_{0}^{1} (1-u^2)^3 \, du \). Now expand \( (1-u^2)^3 = 1 - 3u^2 + 3u^4 - u^6 \). Integrate each term: \[ \int (1 - 3u^2 + 3u^4 - u^6) \, du = u - u^3 + \frac{3}{5}u^5 - \frac{1}{7}u^7 \bigg]_{0}^{1} \].

Step 5: Evaluate the Integrated Expression

Calculate the definite integral by substituting the limits of integration: \( [u - u^3 + \frac{3}{5}u^5 - \frac{1}{7}u^7]_{0}^{1} = (1 - 1 + \frac{3}{5} - \frac{1}{7}) - (0 - 0 + 0 - 0) \).

Step 6: Simplify the Expression

Evaluate the expression: \( 1 - 1 = 0; \frac{3}{5} - \frac{1}{7} = \frac{21}{35} - \frac{5}{35} = \frac{16}{35} \). Thus, the integral value is \( \frac{16}{35} \).

Key Concepts

Trigonometric Integration

Trigonometric integration involves techniques that are particularly useful for integrating functions involving trigonometric expressions. It often takes advantage of trigonometric identities to transform integrals into more workable forms. Common identities used in trigonometric integration include:

• Pythagorean Identities: Such as \( \sin^2 x + \cos^2 x = 1 \)
• Double Angle Identities: \( \sin(2x) = 2\sin(x)\cos(x) \)
• Product-to-Sum Formulas: Useful for integrating products of sines and cosines.

In our example, we used the identity \( \sin^2 y = 1 - \cos^2 y \) to simplify the power of sine into products of powers of cosine which are easier to manage when integrating. This simplification reduces the complexity of trigonometric forms, making them suitable for substitution or other methods.

Integration by Substitution

Integration by substitution is a fundamental technique used to simplify integrals by changing variables. The idea is to transform the integral into a form that is easier to solve. This method is particularly helpful when dealing with compound functions. Here's when you know to use it:

• If the integral is not straightforward and includes a function and its derivative.
• When substituting a variable, it effectively reverses the chain rule, allowing for simpler integration.

As demonstrated, we used the substitution \( u = \cos y \), turning the trigonometric function into a polynomial, transforming the complex trigonometric integral into a relatively simple polynomial integral. The derivative \( du = -\sin y \, dy \) accounted for the extra \( \sin y \) from the original integral and even changed the bounds of integration, which were reversed and simplified into the limits of substitution.

Reduction Formula

The reduction formula is a recursive relationship that expresses a complex integral in terms of a simpler one. It is especially handy for integrals involving powers of trigonometric functions. This method can make what appears to be an impossible integral, possible by reducing the power systematically.Reduction formulas are often derived from integrating by parts or by substitution. They are used repetitively to reduce the power step-by-step.While this problem didn't use a standard reduction formula explicitly, the strategy of substituting sine's odd powers to involve cosine incorporates the basic principle — breaking the integral into simpler parts. This avoids initial indefinite complexity, as noted in solving \( \int \sin^7 y \, dy \) by isolating an odd power as a separate factor.

Odd Powers of Sine

When dealing with odd powers of sine, a nifty trick is to strip one sine factor from the term and rewrite the rest using a trigonometric identity. This is efficient because it allows using substitution effectively.If you have an odd power like \( \sin^7 y \), you can rewrite it as \( \sin^6 y \cdot \sin y \). Now, use the identity \( \sin^2 y = 1 - \cos^2 y \) to change \( \sin^6 y \) into \( (1 - \cos^2 y)^3 \).This decomposition not only makes integration more manageable but also neatly sets up the integral for substitution. Once simplified, substitute \( \cos y \) for \( u \), turning the integral from trigonometric to polynomial, thereby completing the structure needed to solve the integral easily with basic principles.