Use Table 5.6 to evaluate the following indefinite integrals. (e) β« dπ/(81 + 9πΒ²) (Hint: Factor a 9 out of the denominator first.)
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Step 1: Start by factoring a 9 out of the denominator. Rewrite the integral as β« dπ / (9(9 + πΒ²)). This simplifies the denominator and prepares it for further manipulation.
Step 2: Factor out the constant 1/9 from the integral. This gives (1/9) β« dπ / (9 + πΒ²). Constants can be factored out of integrals to simplify calculations.
Step 3: Recognize the standard form of the integral. The denominator (9 + πΒ²) matches the form aΒ² + πΒ², where a = 3. This suggests using the formula β« dπ / (aΒ² + πΒ²) = (1/a) arctan(π/a) + C.
Step 4: Apply the formula. Substitute a = 3 into the formula, resulting in (1/9) * (1/3) arctan(π/3) + C. Combine the constants to simplify the expression further.
Step 5: Write the final simplified integral expression as (1/27) arctan(π/3) + C. This is the indefinite integral of the given function.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Indefinite Integrals
Indefinite integrals represent a family of functions whose derivative is the integrand. They are expressed with the integral sign followed by the function and the differential, and they include a constant of integration, C. Understanding how to evaluate indefinite integrals is crucial for solving problems in calculus, as they provide the antiderivative of a function.
Factoring is a technique used to simplify expressions, making them easier to integrate. In the context of the given problem, factoring a common term from the denominator can transform the integral into a more manageable form. This step is essential for applying integration techniques, such as substitution or recognizing standard integral forms.
Standard integral forms are well-known results that provide the antiderivatives of specific functions. These forms are often found in integral tables and can significantly expedite the integration process. Recognizing when an integral matches a standard form allows for quick evaluation, which is particularly useful in solving complex integrals efficiently.