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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 7.3.49

Evaluate the integrals in Exercises 33–54.
49. ∫ e^(sec πt) sec πt tan πt dt

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1
Identify the integral to solve: \(\int e^{\sec(\pi t)} \sec(\pi t) \tan(\pi t) \, dt\).
Recognize that the integrand contains the function \(e^{\sec(\pi t)}\) multiplied by the derivative of \(\sec(\pi t)\), since the derivative of \(\sec(x)\) is \(\sec(x) \tan(x)\).
Use substitution by letting \(u = \sec(\pi t)\). Then, compute \(du\): since \(\frac{d}{dt} \sec(\pi t) = \pi \sec(\pi t) \tan(\pi t)\), it follows that \(du = \pi \sec(\pi t) \tan(\pi t) \, dt\).
Rewrite the integral in terms of \(u\) and \(du\): solve for \(dt\) in terms of \(du\) and substitute back into the integral to express it fully in terms of \(u\).
Integrate the resulting expression with respect to \(u\), then substitute back \(u = \sec(\pi t)\) to express the answer in terms of \(t\).

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Key Concepts

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

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, differentiating it, and rewriting the integral in terms of this variable. This technique is especially useful when the integral contains a function and its derivative.
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Derivatives of Trigonometric Functions

Understanding the derivatives of trigonometric functions like secant and tangent is crucial. For example, the derivative of sec(x) is sec(x)tan(x), which often appears in integrals involving these functions. Recognizing these derivatives helps in identifying substitution candidates.
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Exponential Functions with Composite Arguments

Exponential functions with composite arguments, such as e^(sec(πt)), require careful handling during integration. The chain rule applies when differentiating or integrating these functions, and recognizing the inner function and its derivative is key to simplifying the integral.
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