Problem 21 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(15-22\). $$ \int_{0}^{\pi / 2} \theta \sqrt{1-\cos 2 \theta} d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sqrt{2} \).

Step 1: Simplify the Integral Expression

First, notice that \( \cos 2\theta = 1 - 2\sin^2\theta \) according to the double angle identity. Therefore, \( 1 - \cos 2\theta = 2\sin^2\theta \). Substituting this into the integral, it becomes \( \int_{0}^{\pi / 2} \theta \sqrt{2\sin^2\theta} \, d\theta = \int_{0}^{\pi / 2} \theta \sqrt{2}\sin\theta \, d\theta \). We can bring \( \sqrt{2} \) outside of the integral as it is a constant, resulting in \( \sqrt{2} \int_{0}^{\pi / 2} \theta \sin\theta \, d\theta \).

Step 2: Use Integration by Parts

The integral \( \int \theta \sin\theta \, d\theta \) can be solved using integration by parts, where \( u = \theta \) and \( dv = \sin\theta \, d\theta \). Then \( du = d\theta \) and \( v = -\cos\theta \). Integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substituting, we get:\[\int \theta \sin\theta \, d\theta = -\theta \cos\theta + \int \cos\theta \, d\theta\]

Step 3: Solve for the Remaining Integral

The remaining integral is \( \int \cos\theta \, d\theta = \sin\theta + C \). Therefore, substituting back, we have:\[-\theta \cos\theta + \sin\theta \]\Applying limits of integration, the expression becomes \(-\theta \cos\theta + \sin\theta |_0^{\pi/2} \).

Step 4: Evaluate the Definite Integral

Evaluate the expression \(-\theta \cos\theta + \sin\theta\) at the limits \(0\) to \(\pi/2\):- At \( \theta = \pi/2 \): \[ -\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = 0 + 1 = 1 \]- At \( \theta = 0 \): \[ -0 \cdot \cos(0) + \sin(0) = 0 \]Thus, the evaluated integral is \(1 - 0 = 1\). Since we factored out \( \sqrt{2} \) earlier, multiply the result by \( \sqrt{2} \) to obtain the final answer.

Step 5: Final Result

Multiply the result by \( \sqrt{2} \), giving final result:\[ \sqrt{2} \times 1 = \sqrt{2} \]

Key Concepts

Integration by Parts

Integration by parts is a powerful technique for solving integrals. It is especially handy when the integral is a product of two functions, like in the example where we have \( \theta \) and \( \sin\theta \). The basic idea is to break this product into parts that are easier to integrate separately.

The formula for integration by parts is derived from the product rule for differentiation, and it is:

• \( \int u \, dv = uv - \int v \, du \)

Here, \( u \) and \( dv \) are parts of your original integral that you choose. For \( \int \theta \sin\theta \, d\theta \), we choose:

• \( u = \theta \), then \( du = d\theta \)
• \( dv = \sin\theta \, d\theta \), then \( v = -\cos\theta \)

This makes integration manageable by turning a complex integral into simpler parts, enabling us to approach the evaluation step by step.

Trigonometric Identities

Trigonometric identities simplify expressions and solve integrals that involve trigonometric functions. In our exercise, we use the double angle identity for cosine to manipulate and simplify the integral.

The identity applied here is the double angle formula for cosine:

• \( \cos 2\theta = 1 - 2\sin^2\theta \)

By rewriting \( \cos 2\theta \) as \( 1 - 2\sin^2\theta \), we transform \( 1 - \cos 2\theta \) into \( 2\sin^2\theta \), allowing us to substitute and simplify the integral to \( 2\sin\theta \) times a constant factor, \( \sqrt{2} \).

These identities are essential for breaking down complex trigonometric expressions into more workable forms.

Double Angle Formulas

Double angle formulas are powerful tools derived from trigonometric identities that simplify the evaluation of integrals involving trigonometric functions. The double angle formula used in this context is:

• \( \cos 2\theta = 1 - 2\sin^2\theta \)

This formula allows us to express a function of \( 2\theta \) in terms of \( \theta \), thus easing the integration process.

In the problem, the expression \( 1 - \cos 2\theta \) simplifies to \( 2\sin^2\theta \). This simplification makes it possible to further break down the integral into simpler components that can be evaluated using standard techniques such as integration by parts. Mastering these formulas can make solving trigonometric integrals much more straightforward.

Definite Integrals Evaluation

Definite integrals give a precise value, representing the area under a curve between two limits. Unlike indefinite integrals, definite integrals are evaluated within given bounds. In the step-by-step solution provided, the process involved solving the integrated function within the constraints from 0 to \( \pi/2 \).

To evaluate a definite integral:

• Solve the indefinite integral first.
• Substitute the upper limit into the solution.
• Subtract the value obtained by substituting the lower limit.

The result is the net area, or total net change, given as a single value. Integrals like \( \int_0^{\pi/2} \theta \sin\theta \, d\theta \) illustrate this well. By computing at bounds \( \theta = \pi/2 \) and \( \theta = 0 \), we find the total area calculated is a numeric result, \( 1 \), before multiplying by \( \sqrt{2} \) as shown in the final step.