Problem 68
Question
Integrals of the form \(\int \sin m x \cos n x d x\) Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}(\cos ((m-n) x)-\cos ((m+n) x))\\\&\sin m x \cos n x=\frac{1}{2}(\sin ((m-n) x)+\sin ((m+n) x))\\\&\cos m x \cos n x=\frac{1}{2}(\cos ((m-n) x)+\cos ((m+n) x))\end{aligned}$$ $$\int \sin 5 x \sin 7 x d x$$
Step-by-Step Solution
Verified Answer
Question: Find the integral of the product of the two trigonometric functions: \(\int \sin 5x \sin 7x dx\).
Answer: The integral is given by \(\int \sin 5x \sin 7x dx = -\frac{1}{4}\sin(-2x) + \frac{1}{24}\sin(12x) + C\), where \(C\) is an arbitrary constant of integration.
1Step 1: Apply identity for the product of sine functions
We will apply the given identity for the product of two sine functions. Given \(\sin 5x \sin 7x\), we have \(m = 5\) and \(n = 7\).
Using the identity:
$$\sin 5x \sin 7x = \frac{1}{2}(\cos (5-7) x - \cos (5+7) x) = \frac{1}{2}(\cos (-2x) - \cos (12x))$$
Now our integral becomes:
$$\int \sin 5x \sin 7x dx = \int \frac{1}{2}(\cos (-2x) - \cos (12x)) dx$$
2Step 2: Integrate the trigonometric terms separately
Split the integral into two parts and integrate each trigonometric term separately:
$$\int \frac{1}{2}(\cos (-2x) - \cos (12x)) dx = \frac{1}{2}\int \cos(-2x) dx - \frac{1}{2} \int \cos (12x) dx$$
For each integral, use the fact that the integral of \(\cos ax\) is \(\frac{1}{a}\sin ax + C\).
Therefore, the first integral is:
$$\frac{1}{2} \int \cos(-2x) dx = -\frac{1}{4}\sin(-2x) + C_1$$
And the second integral is:
$$\frac{1}{2} \int \cos(12x) dx = \frac{1}{24}\sin(12x) + C_2$$
3Step 3: Combine the results and write the final answer
Combine the results from the previous step and write the final answer:
$$\int \sin 5x \sin 7x dx = -\frac{1}{4}\sin(-2x) + \frac{1}{24}\sin(12x) + C$$
Here, \(C\) represents an arbitrary constant of integration (which is the sum of \(C_1\) and \(C_2\)).
Key Concepts
Trigonometric IdentitiesIntegration TechniquesIndefinite Integrals
Trigonometric Identities
Trigonometric identities are mathematical equations that relate various trigonometric functions to one another. In calculus, they are incredibly useful for simplifying integrals that involve products of sine and cosine functions. For instance, when faced with an integral such as \(\int \sin(mx) \cos(nx) dx\), we can use specific identities to rewrite the product into a form that is easier to integrate.
One common identity is \(\sin(mx) \cos(nx) = \frac{1}{2}(\sin((m-n)x) + \sin((m+n)x))\). This identity transforms the product of a sine and cosine into a sum of two sine functions. For products of two sines or two cosines, similar identities exist:
These identities are crucial for turning complex trigonometric expressions into simpler components that can be integrated using standard integration techniques. Understanding and memorizing these identities allow students to approach a wide variety of trigonometric integrals with confidence.
One common identity is \(\sin(mx) \cos(nx) = \frac{1}{2}(\sin((m-n)x) + \sin((m+n)x))\). This identity transforms the product of a sine and cosine into a sum of two sine functions. For products of two sines or two cosines, similar identities exist:
- For sine products: \(\sin(mx) \sin(nx) = \frac{1}{2}(\cos((m-n)x) - \cos((m+n)x))\)
- For cosine products: \(\cos(mx) \cos(nx) = \frac{1}{2}(\cos((m-n)x) + \cos((m+n)x))\)
These identities are crucial for turning complex trigonometric expressions into simpler components that can be integrated using standard integration techniques. Understanding and memorizing these identities allow students to approach a wide variety of trigonometric integrals with confidence.
Integration Techniques
There are several techniques for performing integrations, and knowing when and how to apply them is key to solving calculus problems. The techniques range from simple power rule integration to more advanced methods like integration by parts and partial fractions.
In the example of integrating the product of sine functions, we use a straightforward approach:
Recognizing the structure of an integrand and applying the appropriate technique is essential. Sometimes an integral may require a series of steps or a combination of techniques to arrive at the solution. Practice is the best way to hone these integration skills and become adept at spotting the fastest route to a solution.
In the example of integrating the product of sine functions, we use a straightforward approach:
- First, we apply the relevant trigonometric identity to rewrite the integral into a simpler form.
- Next, we break the integral into terms that can be integrated individually.
- Finally, we integrate each term using basic integration rules. In the case of \(\cos(ax)\), the integral is \(\frac{1}{a}\sin(ax) + C\), where \(C\) represents the integration constant.
Recognizing the structure of an integrand and applying the appropriate technique is essential. Sometimes an integral may require a series of steps or a combination of techniques to arrive at the solution. Practice is the best way to hone these integration skills and become adept at spotting the fastest route to a solution.
Indefinite Integrals
An indefinite integral, also known as an antiderivative, is a function that reverses the process of differentiation. When we calculate an indefinite integral, we're finding a function whose derivative gives the original function. An indefinite integral is represented as \(\int f(x) dx\), and it includes a constant of integration \(C\) because differentiation doesn't reveal constant terms.
For example, in the integral of a cosine function \(\int \cos(ax) dx\), the result is \(\frac{1}{a}\sin(ax) + C\). Here, \(C\) symbolizes that there are an infinite number of antiderivatives, since adding any constant to a function does not change its derivative.
The technique for finding an indefinite integral can vary depending on the function we are integrating. It's worth noting that not all functions have indefinite integrals that can be expressed in terms of elementary functions, and this is where numerical integration techniques or approximations may come into play. A firm grasp of indefinite integrals is pivotal for problem-solving in calculus, as it forms the basis for understanding the areas under curves and the accumulated quantities represented by integrals.
For example, in the integral of a cosine function \(\int \cos(ax) dx\), the result is \(\frac{1}{a}\sin(ax) + C\). Here, \(C\) symbolizes that there are an infinite number of antiderivatives, since adding any constant to a function does not change its derivative.
The technique for finding an indefinite integral can vary depending on the function we are integrating. It's worth noting that not all functions have indefinite integrals that can be expressed in terms of elementary functions, and this is where numerical integration techniques or approximations may come into play. A firm grasp of indefinite integrals is pivotal for problem-solving in calculus, as it forms the basis for understanding the areas under curves and the accumulated quantities represented by integrals.
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