Question

67-70. Integrals of the form \(\int \sin m x\) cos nux dx Use the following three identities to evaluate the given integrals. $$ \begin{aligned} \sin m x \sin n x &=\frac{1}{2}(\cos ((m-n) x)-\cos ((m+n) x)) \\ \sin m x \cos n x &=\frac{1}{2}(\sin ((m-n) x)+\sin ((m+n) x)) \\\ \cos m x \cos n x &=\frac{1}{2}(\cos ((m-n) x)+\cos ((m+n) x)) \end{aligned} $$ $$\int \sin 3 x \sin 2 x \, d x$$

Short answer

After applying the sine product identity and integrating each term separately, we find that the integral of sin(3x)sin(2x) with respect to x is given by: $$\int \sin 3x \sin 2x \, dx = \frac{1}{2}\sin x - \frac{1}{10}\sin 5x + C.$$

Step-by-step solution

Use the sine product identity

To begin, we will use the given identity for the product of sines: $$ \sin mx \sin nx = \frac{1}{2}(\cos((m-n)x) - \cos((m+n)x)). $$ In this case, we have \(m=3\) and \(n=2\). Apply the identity to the given function inside the integral: $$ \sin 3x \sin 2x = \frac{1}{2}(\cos((3-2)x) - \cos((3+2)x)) = \frac{1}{2}(\cos x - \cos 5x). $$

Substitute the expression inside the integral

Replace the product of sines in the integral with the expression we found in step 1: $$ \int \sin 3x \sin 2x \, dx = \int \frac{1}{2}(\cos x - \cos 5x) \, dx. $$

Distribute the constant and separate the integral

Distribute the \(\frac{1}{2}\) and separate the integral into two parts: $$ \int \frac{1}{2}(\cos x - \cos 5x) \, dx = \frac{1}{2}\int \cos x \, dx - \frac{1}{2}\int \cos 5x \, dx. $$

Integrate each term separately

We can now individually integrate each term in the expression: $$ \frac{1}{2}\int \cos x \, dx = \frac{1}{2}\sin x + C_1, $$ and $$ \frac{1}{2}\int \cos 5x \, dx = \frac{1}{10}\sin 5x + C_2. $$

Combine the results and finalize the solution

Combine the results from Step 4 and write the final answer: $$ \int \sin 3x \sin 2x \, dx = \frac{1}{2}\sin x - \frac{1}{10}\sin 5x + C. $$

Key concepts

Integration Techniques

Understanding integration techniques is crucial when facing a complex integral. It involves recognizing patterns and applying specific methods to simplify and compute the integral effectively. To solve the integral \(\int \sin 3x \sin 2x\, dx\), we employ a technique that involves using trigonometric identities to transform the product of trigonometric functions into a sum or difference that is easier to integrate.

The first step is to identify and apply the product-to-sum identity specifically tailored for the sine functions. This identity allows the integral to be broken down into simpler parts that can be easily managed. Once the identity is applied and the integral is expressed as a sum of cosines, the constant factor is distributed, and the integral is separated into more manageable parts.

This process of separating and simplifying is an essential skill in integrating more complex expressions, and the ability to spot opportunities to apply these trigonometric identities can significantly streamline the integration process.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all the input values for which the functions are defined. They are fundamental in simplifying expressions and solving equations involving trigonometric functions.

In the context of integration, applying the right trigonometric identity can transform an intimidating integral into a more straightforward problem. For instance, the product-to-sum identities used in the integral of \(\int \sin 3x \sin 2x\, dx\) allow us to rewrite the product of sine functions as a sum and difference of cosine functions. This is done by utilizing the identity
\[\sin mx \sin nx = \frac{1}{2}(\cos((m-n)x) - \cos((m+n)x))\]
where \(m\) and \(n\) are constants that represent the frequencies of the sine functions being multiplied. The powerful transformation drastically simplifies the integration task and exemplifies the essential role that trigonometric identities play in calculus.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, represent the general form of the integral without specified limits. Solving an indefinite integral means finding all functions that, once differentiated, result in the original function.

After applying the product-to-sum trigonometric identity and separating the integral into simpler components, the next step is to find the indefinite integrals of each part. In our example, the separate integrals of \(\cos x\) and \(\cos 5x\) both require the basic knowledge of the antiderivative of cosine, which is the sine function.

Finding indefinite integrals often involves recognizing the derivative forms present within the integral. Due to the constant integration by all mathematical operations present in the expression, it is crucial to keep track of constants and apply any necessary adjustments, as seen with the coefficient in front of the integral of \(\cos 5x\). At the end of the process, the constant of integration, \(C\), is added to represent the family of all antiderivatives.