Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-62, use the Substitution Rule for Definite Integrals to evaluate each definite integral. \(\int_{0}^{\pi / 2} \sin ^{2} 3 x \cos 3 x d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{-1}{9}\).

Step 1: Identify the Substitution

We notice that the integrand is of the form \(\sin^2(3x)\cos(3x)\). We can set \(u = \sin(3x)\) as a substitution, which simplifies the integrand when differentiating. The differential \(du = 3\cos(3x)\,dx\), so \(dx = \frac{du}{3\cos(3x)}\).

Step 2: Change the Bounds of Integration

When \(x = 0\), \(u = \sin(3\times0) = 0\). When \(x = \frac{\pi}{2}\), \(u = \sin(\frac{3\pi}{2}) = -1\). So, the new bounds for \(u\) are from 0 to -1.

Step 3: Substitute and Simplify the Integral

Using the substitution, the integral becomes:\[\int_{0}^{-1} u^2 \frac{du}{3} = \frac{1}{3} \int_{0}^{-1} u^2 \, du\]Recall that when we substituted, the \(\cos(3x)\) terms canceled out.

Step 4: Evaluate the Integral

Now, compute the integral:\[\frac{1}{3} \int_{0}^{-1} u^2 \, du = \frac{1}{3} \left[ \frac{u^3}{3} \right]_{0}^{-1} = \frac{1}{9} \left[ (-1)^3 - 0^3 \right]\]This results in the evaluation:\[\frac{1}{9}(-1)\].

Step 5: Calculate and Conclude

The final calculation results in:\[\frac{-1}{9}\]Thus, the value of the definite integral \( \int_{0}^{\pi / 2} \sin^2 3x \cos 3x \, dx \) is \(\frac{-1}{9}\).

Key Concepts

Definite Integrals

Definite integrals allow us to calculate the net area under a curve over a specific interval. Unlike indefinite integrals, definite integrals focus on finding the total accumulation from one point to another along the x-axis. In the original problem, we see a definite integral \[ \int_{0}^{\pi / 2} \sin^2 3x \cos 3x \, dx \]The limits of integration, 0 to \( \pi/2 \), denote the interval over which we are measuring the area. By using the substitution rule, we replace complex expressions with simpler functions. This makes it easier to integrate, especially for expressions involving trigonometric functions.

• The integral is bounded, meaning it provides the net area under the curve between two points.
• Definite integrals can result in a specific numerical value, which represents total accumulation or net change.

In our example, the steps of substitution help in converting a complicated trigonometric integral into a polynomial one.

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals that contain trigonometric expressions. By strategically replacing trigonometric identities with variables, we can transform the integral into a form that is easier to evaluate.
For the given problem, we used the substitution \(u = \sin(3x)\), which simplifies the original expression. This substitution turns the expression \(\sin^2(3x)\cos(3x)\) into \(u^2\), resulting in a polynomial integral form. This transformation hinges on identifying supportive substitutions that effectively reduce the complexity of the integral.

• Choose substitutions that simplify the integrand, often by equating them to trigonometric functions.
• Convert differentials accordingly, as seen in our problem where \(dx = \frac{du}{3\cos(3x)}\).
• Ensure all trigonometric components are accounted for when substituting values.

Using trigonometric substitution adequately addresses complex trigonometric integrals by transforming them into simpler algebraic expressions.

Integral Bounds

Integral bounds refer to the limits between which the definite integral is evaluated. Changing the variable of integration through substitution also requires adjusting these bounds to reflect the new variable
In our exercise, when substituting \(u = \sin(3x)\), we needed to compute what happens to the bounds of integration:
- Originally, when \(x = 0\), \(u = \sin(0) = 0\).- When \(x = \pi/2\), \(u = \sin(3\pi/2) = -1\).These transformed bounds allow us to evaluate the integral in the new variable \(u\).

• Always substitute bounds of integration to match the new variable.
• Evaluate the transformed integral within the new limits to find the correct area under the curve.

Handling integral bounds correctly is crucial for solving any definite integral using substitution, ensuring accurate results for the net area calculation.