Question

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. \(\int_{0}^{\pi} \cos (99 x) \sin (101 x) d x\)

Short answer

The value of the definite integral is 0.

Step-by-step solution

Use Trigonometric Identity

We start by using the product-to-sum identity for trigonometric functions: \[\cos A \sin B = \frac{1}{2} [\sin(A + B) - \sin(A - B)]\]Applying this identity to our integral, let \( A = 99x \) and \( B = 101x \). We have:\[\cos(99x)\sin(101x) = \frac{1}{2} [\sin(200x) - \sin(2x)]\]

Express the Integral with the Identity

Substitute the expression from Step 1 into the integral:\[\int_0^\pi \cos(99x)\sin(101x) \, dx = \int_0^\pi \frac{1}{2} [\sin(200x) - \sin(2x)] \, dx\]Factor out the \( \frac{1}{2} \):\[= \frac{1}{2} \left( \int_0^\pi \sin(200x) \, dx - \int_0^\pi \sin(2x) \, dx \right)\]

Integrate Each Term Separately

Integrate each sine function separately:1. For \( \int_0^\pi \sin(200x) \, dx \): - The antiderivative is \(-\frac{1}{200} \cos(200x)\). - Evaluate from 0 to \( \pi \): - \(-\frac{1}{200} [\cos(200\pi) - \cos(0)] = -\frac{1}{200} [1 - 1] = 0\).2. For \( \int_0^\pi \sin(2x) \, dx \): - The antiderivative is \(-\frac{1}{2} \cos(2x)\). - Evaluate from 0 to \( \pi \): - \(-\frac{1}{2} [\cos(2\pi) - \cos(0)] = -\frac{1}{2} [1 - 1] = 0\).Both integrals evaluate to zero.

Calculate the Final Result

Substitute the evaluated integrals back into the factored expression:\[\frac{1}{2} \left( 0 - 0 \right) = 0\]Thus, the value of the definite integral is 0.

Key concepts

Trigonometric Identities

Trigonometric identities are mathematical relationships involving trigonometric functions like sine, cosine, and tangent. These identities help simplify complex expressions and solve integrals or equations that involve trigonometric functions. In this exercise, the product-to-sum identity was used, which is one of the most helpful identities in trigonometry. This particular identity allows you to rewrite a product of sine and cosine as a sum or difference of two other trigonometric functions.
Consider the formula: \(\cos A \sin B = \frac{1}{2} [\sin(A + B) - \sin(A - B)]\). This conversion is crucial when evaluating integrals involving the product of sine and cosine terms, as it makes integration possible by transforming the product into a sum.

• The identity helps in reducing complex expressions into simpler ones.
• It is especially useful in calculus for solving definite and indefinite integrals involving trigonometric functions.

Integral Calculus

Integral calculus is a fundamental part of mathematics focusing on finding the integral of a function. It deals with accumulation quantities, such as areas under curves, volumes, and other concepts. In this exercise, we focus on evaluating definite integrals, which are integrals with specified upper and lower bounds.
A definite integral is represented by the symbol \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the bounds. When you compute a definite integral of a function over a specified interval, you essentially find the net area under the curve of the function between these two bounds.

• Definite integrals can represent actual physical quantities like distances, areas, or volumes.
• To solve these integrals, you may need to use antiderivatives and evaluate at the specific bounds.

In this example, we transformed the integral \(\int_{0}^{\pi} \cos(99x) \sin(101x) \ dx\) into more easily integrable parts, using trigonometric identities. This made evaluation straightforward as the resulting component-wise integrals evaluated to zero.

Product-to-Sum Formulas

The Product-to-Sum formulas in trigonometry allow transforming the product of two trig functions into a sum or difference of trig functions. This transformation is immensely useful, especially in integral calculus, where dealing with sums is typically easier than dealing with products.
For example, the formula for the product of cosine and sine used in the exercise is \(\cos A \sin B = \frac{1}{2} [\sin(A + B) - \sin(A - B)]\). By using this formula, complex integrals can be broken down into simpler forms, as was done in the exercise for the definite integral \(\int_{0}^{\pi} \cos(99x) \sin(101x) \ dx\).

• These formulas simplify integration by breaking products into accessible sums and differences.
• They are derived from the properties of trigonometric functions and their symmetry.

When evaluating the integral in this exercise, the use of the product-to-sum formula greatly simplified the task, allowing us to end up with components that were zero, simplifying the overall process to find the final result.