Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

, use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{\pi / 6} \sin ^{3} \theta \cos \theta d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{64} \).

Step 1: Identify the substitution

To apply the substitution rule, we first need to find a substitution that will simplify the integral. Observe that the integrand includes both \( \sin^3 \theta \) and \( \cos \theta \), suggesting a substitution involving \( \sin \theta \). Let \( u = \sin \theta \), then \( du = \cos \theta \, d\theta \).

Step 2: Change the limits of integration

When we change variables, we also need to adjust the limits of integration to reflect the substitution. Originally, \( \theta \) ranges from \( 0 \) to \( \pi/6 \). Substitute these values into \( u = \sin \theta \) to find the new limits:- When \( \theta = 0 \), \( u = \sin 0 = 0 \).- When \( \theta = \pi/6 \), \( u = \sin(\pi/6) = 1/2 \).Thus, the new limits are 0 to 1/2.

Step 3: Rewrite the integrand with substitution

Substitute \( u = \sin \theta \) and \( du = \cos \theta \, d\theta \) into the integral:\[ \int_{0}^{\pi/6} \sin^3 \theta \cos \theta \, d\theta = \int_{0}^{1/2} u^3 \, du \]

Step 4: Evaluate the new integral

Evaluate the integral \( \int_{0}^{1/2} u^3 \, du \). Use the power rule for integration:\[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C \]Substitute \( n = 3 \):\[ \int u^3 \, du = \frac{u^4}{4} \]Now, apply the limits from 0 to 1/2:\[ \left[ \frac{u^4}{4} \right]_{0}^{1/2} = \frac{(1/2)^4}{4} - \frac{(0)^4}{4} \]

Step 5: Simplify and calculate the result

Calculate the expression:\[ \frac{(1/2)^4}{4} = \frac{1}{16}/4 = \frac{1}{64} \]Thus, the definite integral evaluates to \( \frac{1}{64} \).

Key Concepts

Definite Integral

A definite integral is an essential concept in calculus that calculates the accumulation of quantities, such as area, volume, or other physical properties, between two limits. In this context, the definite integral \( \int_{a}^{b} f(x) \, dx \) represents the area under the curve \( f(x) \) from \( x = a \) to \( x = b \) on the x-axis. The limits \( a \) and \( b \) are essential as they define the interval over which the function is integrated.

• It provides a way to calculate the net accumulation or loss within particular bounds – the limits of integration.
• The function \( f(x) \) is the integrand, which is integrated over the interval \( [a, b] \).

In our exercise, the integrand is \( \sin^3 \theta \cos \theta \), encompassing trigonometric functions that require specific integration techniques, like substitution, to effectively evaluate the integral from \( \theta = 0 \) to \( \theta = \frac{\pi}{6} \).

Trigonometric Substitution

Trigonometric substitution is a clever technique used to simplify integrals containing trigonometric functions. It often turns a complex problem into a more straightforward polynomial integral, which is much easier to solve. Here's how it works:The idea is to identify a suitable substitution that simplifies the integrand. In our example, the substitution \( u = \sin \theta \) is used. This choice is because our integrand includes \( \sin^3 \theta \) and \( \cos \theta \), and substituting \( u = \sin \theta \) converts the integral's form.

• After the substitution, the differential \( du = \cos \theta \, d\theta \) aligns perfectly with the integral's remaining term \( \cos \theta \), greatly simplifying the process.
• The integral \( \int_{0}^{\pi/6} \sin^3 \theta \cos \theta \, d\theta \) becomes \( \int_{0}^{1/2} u^3 \, du \), a straightforward polynomial integration challenge.

By changing the variable through trigonometric substitution, complex integrals become easier to handle, facilitating the evaluation process.

Limits of Integration

The limits of integration are crucial in determining the region of accumulation for a definite integral. When variables change through substitution, the limits of integration must also change to reflect the new variable. This ensures the results precisely cover the same interval as the original integral.In our exercise, the original limits for the variable \( \theta \) were \( 0 \) to \( \frac{\pi}{6} \). After applying the substitution \( u = \sin \theta \), these limits were recalculated:

• When \( \theta = 0 \), \( u = \sin 0 = 0 \).
• When \( \theta = \frac{\pi}{6} \), \( u = \sin(\frac{\pi}{6}) = \frac{1}{2} \).

Thus, the new limits for \( u \) are \( 0 \) to \( \frac{1}{2} \).Leaving out the adjustment of limits can lead to incorrect results, as the range over which the function is integrated must remain consistent with the substitution. It ensures that the area or quantity measured is directly comparable to the original problem's bounds.