Question

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 5 x \sin 7 x d x$$

Short answer

Question: Evaluate the integral $$\int \sin 5x \sin 7x\ dx$$. Answer: $$\frac{-1}{4}\sin(-2x) - \frac{1}{24}\sin(12x) + C$$

Step-by-step solution

Apply trigonometric identity

Replace the product of the sine functions in the integral using the given identity. Our integral will become: $$\int \frac{1}{2}[\cos((5-7)x) - \cos((5+7)x)]\ dx$$

Simplify the integral

Now simplify the integral by dealing with the constants: $$\frac{1}{2} \int [\cos(-2x) - \cos(12x)]\ dx$$

Split the integral

We can split the integral into two separate integrals: $$\frac{1}{2} \left[\int \cos(-2x)\ dx - \int \cos(12x)\ dx\right]$$

Integrate separately

Integrate each part separately. The integral of \(\cos(ax)\) is \(\frac{1}{a}\sin(ax)\). Therefore, we have: $$\frac{1}{2} \left[\frac{1}{-2}\sin(-2x) - \frac{1}{12}\sin(12x)\right] + C$$

Simplify and write the final answer

Now simplify the result to obtain the final answer: $$\frac{-1}{4}\sin(-2x) - \frac{1}{24}\sin(12x) + C$$

Key concepts

Trigonometric Identities

Trigonometric identities are essential tools in calculus, especially when dealing with integrals involving trigonometric functions. They allow us to simplify complex expressions using certain established formulas. In this exercise, we specifically use product-to-sum formulas, which transform products of trigonometric functions into sums or differences.

For example, the identity \( \sin mx \sin nx = \frac{1}{2}[\cos((m-n)x) - \cos((m+n)x)] \) is key for evaluating integrals like \( \int \sin 5x \sin 7x \, dx \). This identity helps us convert the product of two sine functions into a more manageable form involving cosine. Transformations like these are crucial because they simplify the process of evaluating integrals into basic integrals of sine and cosine that we know how to solve.

Understanding and remembering these identities can drastically streamline solving complex trigonometric integrals in calculus.

Definite and Indefinite Integrals

In calculus, integrals are classified into two main types: definite and indefinite integrals. Definite integrals provide exact values, representing the area under a curve within specified limits, while indefinite integrals represent a family of functions, including a constant of integration \(C\).

Since the exercise at hand involves an indefinite integral, its result is a function plus the integration constant. This constant \(C\) acknowledges that there could be multiple antiderivatives for any given function.

The indefinite integral of a function generally has the form \( \int f(x) \, dx = F(x) + C \), where \(F(x)\) is an antiderivative of \(f(x)\). For trigonometric integrals, recognizing the basic antiderivatives of sine and cosine is crucial. For instance, \( \int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C \). Teaching students to master these fundamentals helps them handle more advanced integration problems efficiently.

Product-to-Sum Formulas

Product-to-sum formulas are a type of trigonometric identity used to simplify expressions involving the products of trigonometric functions. These formulas are invaluable in integrals because they facilitate the transformation of these products into a sum or difference of simpler trigonometric terms, which are much easier to integrate.

In this exercise, we used the product-to-sum formula for sine functions: \( \sin m x \sin n x=\frac{1}{2}[\cos ((m-n)x) - \cos ((m+n)x)] \). This identity allowed us to convert the integral \( \int \sin 5x \sin 7x \, dx \) into the form \( \frac{1}{2}\int[\cos(-2x) - \cos(12x)] \, dx \), making it simpler to evaluate.

Recognizing when and how to use these formulas is fundamental in trigonometry-related calculus problems. They not only offer a methodical approach to solving problems but also enhance understanding of the relationships between trigonometric functions.