Problem 27
Question
Calculate. (If you run out of ideas, use the examples as models.) $$\int \sin 5 x \sin 2 x d x$$.
Step-by-Step Solution
Verified Answer
The short answer for the integral of \(\sin{5x} \sin{2x}\) is:
\[\int \sin{5x} \sin{2x} \, dx = \frac{1}{6} \sin{3x} - \frac{1}{14}\sin{7x} + C\]
1Step 1: Apply the product-to-sum identity
First, apply the product-to-sum identity to convert the product of the two sine functions, \(\sin{5x} \sin{2x}\), into a sum:
\[\sin{5x} \sin{2x} = \frac{1}{2} (\cos{(5x - 2x)} - \cos{(5x + 2x)})\]
Thus, we get:
\[\sin{5x} \sin{2x} = \frac{1}{2} (\cos{3x} - \cos{7x})\]
Now we can rewrite the integral as:
\[\int \sin{5x} \sin{2x} \, dx = \int \frac{1}{2} (\cos{3x} - \cos{7x}) \, dx\]
2Step 2: Integrate the simplified expression
Next, we will integrate the simplified expression term by term:
\[\int \frac{1}{2} (\cos{3x} - \cos{7x}) \, dx = \frac{1}{2} \int (\cos{3x} - \cos{7x}) \, dx\]
The antiderivative of \(\cos{nx}\) is \(\frac{1}{n} \sin{nx}\) modulated with a constant term, so we have:
\[\frac{1}{2} \int (\cos{3x} - \cos{7x}) \, dx = \frac{1}{2} (\frac{1}{3} \sin{3x} - \frac{1}{7}\sin{7x}) + C\]
3Step 3: Write the final answer
Finally, write out the answer for the integral in a simplified form:
\[\int \sin{5x} \sin{2x} \, dx = \frac{1}{6} \sin{3x} - \frac{1}{14}\sin{7x} + C\]
Where \(C\) is the constant of integration.
Key Concepts
Product-to-sum IdentitiesAntiderivativesIntegration Methods
Product-to-sum Identities
When integrating trigonometric functions, particularly the product of two trigonometric functions, a helpful tool in our arsenal is the product-to-sum identities. These identities transform the product of two sines, two cosines, or a sine and a cosine into a sum (or difference) of two trigonometric functions. For example, the product-to-sum identities for sine functions, like the one used in our exercise, states:
\[\begin{equation}\sin A \times \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\end{equation}\]
These identities simplify integration because it is generally easier to integrate a sum of functions rather than their product. In the given example, we converted \[\begin{equation}\sin 5x \times \sin 2x\end{equation}\]to \[\begin{equation}\frac{1}{2}(\cos 3x - \cos 7x)\end{equation}\], which was significantly more straightforward to integrate.
\[\begin{equation}\sin A \times \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\end{equation}\]
These identities simplify integration because it is generally easier to integrate a sum of functions rather than their product. In the given example, we converted \[\begin{equation}\sin 5x \times \sin 2x\end{equation}\]to \[\begin{equation}\frac{1}{2}(\cos 3x - \cos 7x)\end{equation}\], which was significantly more straightforward to integrate.
Antiderivatives
The concept of antiderivatives is fundamental to the process of integration. An antiderivative of a function f(x) is a function F(x) such that
\[\begin{equation}F'(x) = f(x)\end{equation}\]
That is, the derivative of F(x) with respect to x gives us back our original function f(x). This process is essentially 'undoing' differentiation. For trigonometric functions, like cosines and sines in our example, their antiderivatives are well-known. For instance, the antiderivative of \[\begin{equation}\cos(nx)\end{equation}\]is \[\begin{equation}\frac{1}{n}\sin(nx)\end{equation}\]plus a constant of integration, denoted by C. This constant represents the infinite number of antiderivatives a function can have since differentiation eliminates any constant terms. It is crucial when integrating to always include this constant to represent the general solution to the integral.
\[\begin{equation}F'(x) = f(x)\end{equation}\]
That is, the derivative of F(x) with respect to x gives us back our original function f(x). This process is essentially 'undoing' differentiation. For trigonometric functions, like cosines and sines in our example, their antiderivatives are well-known. For instance, the antiderivative of \[\begin{equation}\cos(nx)\end{equation}\]is \[\begin{equation}\frac{1}{n}\sin(nx)\end{equation}\]plus a constant of integration, denoted by C. This constant represents the infinite number of antiderivatives a function can have since differentiation eliminates any constant terms. It is crucial when integrating to always include this constant to represent the general solution to the integral.
Integration Methods
Finding the antiderivative can be approached through various integration methods. The method to choose often depends on the form of the function we are dealing with. In the given problem, after employing the product-to-sum identities, we were left with an easy-to-handle function that needed a straightforward application of the power and trigonometric rules of integration.
Integrating term by term, we applied the known antiderivatives of the cosine function. In other scenarios, one might use integration by parts, substitution, or for more challenging integrals, partial fractions, and trigonometric substitution. Being familiar with a wide array of methods equips us to tackle complex integrals more efficiently.
It's important to recognize the type of function we are working with and decide the best method to proceed, sometimes combining several techniques to achieve the result.
Integrating term by term, we applied the known antiderivatives of the cosine function. In other scenarios, one might use integration by parts, substitution, or for more challenging integrals, partial fractions, and trigonometric substitution. Being familiar with a wide array of methods equips us to tackle complex integrals more efficiently.
It's important to recognize the type of function we are working with and decide the best method to proceed, sometimes combining several techniques to achieve the result.
Other exercises in this chapter
Problem 26
Calculate. $$\int e^{x} \tan e^{x} d x$$
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Calculate. $$\int \frac{x-3}{x^{3}+x^{2}} d x$$
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Calculate. $$\int \frac{d x}{e^{x} \sqrt{e^{2 x}-9}}$$.
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Calculate. $$\int \frac{1}{\sin x+\tan x} d x$$
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