Question

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. \(\int_{0}^{\pi} \sin 3 x \sin 5 x d x\)

Short answer

The definite integral evaluates to 0.

Step-by-step solution

Recognize the Integral Type

The integral \( \int_{0}^{\pi} \sin 3x \sin 5x \; dx \) can be solved using trigonometric identities for the product of sine functions. The integral involves two sine functions multiplied together.

Use Product-to-Sum Formulas

Apply the product-to-sum identities to simplify \( \sin 3x \sin 5x \). The identity is given by: \[ \sin A \sin B = \frac{1}{2} \left[ \cos(A-B) - \cos(A+B) \right]. \]Using this, \[ \sin 3x \sin 5x = \frac{1}{2} \left[ \cos(5x - 3x) - \cos(5x + 3x) \right] = \frac{1}{2} [ \cos 2x - \cos 8x ].\]

Substitute into the Integral

Substitute the expression found in Step 2 back into the integral:\[\int_{0}^{\pi} \sin 3x \sin 5x \; dx = \frac{1}{2} \int_{0}^{\pi} (\cos 2x - \cos 8x) \; dx.\]

Integrate Each Term

Integrate each cosine term separately. The integral of \( \cos kx \) is:\[ \int \cos kx \; dx = \frac{1}{k} \sin kx + C.\]Therefore, \[\int \cos 2x \; dx = \frac{1}{2} \sin 2x,\]\[\int \cos 8x \; dx = \frac{1}{8} \sin 8x.\]

Evaluate Definite Integrals

Find the definite integrals:\[\frac{1}{2} \int_{0}^{\pi} \cos 2x \; dx = \frac{1}{2} \left[\frac{1}{2} \sin 2x \right]_{0}^{\pi} = \frac{1}{4} [\sin 2\pi - \sin 0] = \frac{1}{4} [0 - 0] = 0,\]\[\frac{1}{2} \int_{0}^{\pi} \cos 8x \; dx = \frac{1}{2} \left[\frac{1}{8} \sin 8x \right]_{0}^{\pi} = \frac{1}{16} [\sin 8\pi - \sin 0] = \frac{1}{16} [0 - 0] = 0.\]

Combine the Results

Combine the results of the definite integrals:\[\frac{1}{2} (0 - 0) = 0.\]Thus, the integral evaluates to 0.

Key concepts

Trigonometric Identities

To evaluate definite integrals involving trigonometric functions, it is often essential to use trigonometric identities. These identities are mathematical equations that express relationships between trigonometric functions and are crucial in simplifying integrals.

For instance, in the integral \( \int_{0}^{\pi} \sin 3x \sin 5x \, dx \), the problem involves two sine functions. By recognizing the type of functions involved, we can determine the appropriate identity to apply. Trigonometric identities enable us to transform complex expressions into simpler forms that are easier to work with.

This approach reduces computation time and complexity, allowing for a more straightforward solution to otherwise challenging integrals.

Product-to-Sum Formulas

When tackling integrals with products of trigonometric functions, the product-to-sum formulas come in handy. These formulas convert products of sine and cosine functions into sums or differences, facilitating easier integration.

The relevant identity for the integral \( \sin A \sin B \) is: \[ \sin A \sin B = \frac{1}{2} \left[ \cos(A-B) - \cos(A+B) \right]. \] In our example, applying this formula to \( \sin 3x \sin 5x \) simplifies it to \( \frac{1}{2} \left[ \cos 2x - \cos 8x \right] \).

This transformation is valuable because integrating basic cosine functions is more straightforward as opposed to directly integrating a product of sines.

Cosine Integration

Integrating cosine terms is generally straightforward thanks to standard integration formulas. For any cosine function, \( \cos kx \), the integral is given by: \[ \int \cos kx \, dx = \frac{1}{k} \sin kx + C. \]

Using this formula for our example, we find that: - \( \int \cos 2x \, dx = \frac{1}{2} \sin 2x \)- \( \int \cos 8x \, dx = \frac{1}{8} \sin 8x \)

This is helpful because turning a cosine function into its sine counterpart simplifies the integration process, making it more efficient to evaluate within definite limits.

Integral Evaluation

Evaluating definite integrals involves computing the antiderivative and then determining the result over specified limits. Once you have the antiderivative of the function, use the fundamental theorem of calculus to evaluate the definite integral.

In our problem, after integrating each term, we calculate: \[ \frac{1}{2} \int_{0}^{\pi} \cos 2x \, dx = \frac{1}{4} [\sin 2x]_{0}^{\pi} = 0, \]\[ \frac{1}{2} \int_{0}^{\pi} \cos 8x \, dx = \frac{1}{16} [\sin 8x]_{0}^{\pi} = 0. \]

This means that the areas under the curve cancel out over the interval, resulting in a zero. It demonstrates how symmetrical properties and periodicity of trigonometric functions often lead to zero results in definite integrals.

Integration Techniques

Various integration techniques simplify complex integral evaluations. Recognizing which technique to use is crucial for efficiently solving integrals.

Some common techniques include:

• Substitution: Used to simplify the integrand by changing the variable.
• Integration by parts: Useful for products of functions.
• Trigonometric substitution: Effective for integrals involving the square roots of quadratic expressions.

In the given problem, the product-to-sum identity serves as a powerful technique, transforming the product of sines into manageable cosine terms. This showcases how selecting the right technique can make the solution path significantly more straightforward.