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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.PE.39

Evaluate the integrals in Exercises 31–78.
39. ∫(from 0 to π)tan(x/3)dx

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1
Identify the integral to be evaluated: \(\int_0^{\pi} \tan\left(\frac{x}{3}\right) \, dx\).
Use a substitution to simplify the integral. Let \(u = \frac{x}{3}\), which implies \(x = 3u\) and \(dx = 3 \, du\).
Change the limits of integration according to the substitution: when \(x = 0\), then \(u = 0\); when \(x = \pi\), then \(u = \frac{\pi}{3}\).
Rewrite the integral in terms of \(u\): \(\int_0^{\pi} \tan\left(\frac{x}{3}\right) \, dx = \int_0^{\frac{\pi}{3}} \tan(u) \cdot 3 \, du = 3 \int_0^{\frac{\pi}{3}} \tan(u) \, du\).
Recall the integral formula for tangent: \(\int \tan(u) \, du = -\ln|\cos(u)| + C\). Use this to express the integral and then apply the limits from \(0\) to \(\frac{\pi}{3}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integral

A definite integral calculates the net area under a curve between two specified limits. It is represented as ∫ from a to b of f(x) dx, where a and b are the lower and upper bounds. Evaluating definite integrals often involves finding an antiderivative and then applying the Fundamental Theorem of Calculus.
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Integration of Trigonometric Functions

Integrating trigonometric functions like tangent requires knowledge of their antiderivatives. For example, the integral of tan(x) is -ln|cos(x)| + C. Recognizing these standard forms helps simplify the integration process, especially when the argument of the function is a linear expression.
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Substitution Method

The substitution method simplifies integrals by changing variables to transform the integral into a more familiar form. For integrals like ∫tan(x/3) dx, setting u = x/3 helps adjust the limits and integrand accordingly, making the integral easier to evaluate.
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