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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblem 7.1.54
Chapter 7, Problem 7.1.54

29–62. Integrals Evaluate the following integrals. Include absolute values only when needed.


∫₀^{π} 2^{sin x} · cos x dx

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1
Recognize that the integral is of the form \(\int_0^{\pi} 2^{\sin x} \cdot \cos x \, dx\), where the integrand involves an exponential function with a trigonometric exponent multiplied by \(\cos x\).
Use substitution to simplify the integral. Let \(u = \sin x\). Then, compute the differential \(du = \cos x \, dx\).
Rewrite the integral in terms of \(u\): since \(du = \cos x \, dx\), the integral becomes \(\int_{u=\sin 0}^{u=\sin \pi} 2^u \, du\).
Evaluate the new limits of integration: \(\sin 0 = 0\) and \(\sin \pi = 0\), so the integral becomes \(\int_0^0 2^u \, du\).
Notice that the limits of integration are the same, which implies the value of the integral is zero without further calculation.

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Key Concepts

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

Definite Integrals

A definite integral calculates the net area under a curve between two limits, here from 0 to π. It involves evaluating the integral of a function over a specified interval, resulting in a numerical value that represents accumulation or total change.
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Definition of the Definite Integral

Integration by Substitution

Integration by substitution simplifies integrals by changing variables to transform the integral into a more manageable form. It is especially useful when the integrand contains a function and its derivative, allowing easier evaluation.
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Substitution With an Extra Variable

Properties of Exponential Functions with Variable Exponents

Functions like 2^(sin x) involve an exponential with a variable exponent. Understanding how to handle such expressions, especially when combined with trigonometric functions, is key to setting up substitutions and simplifying the integral.
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Properties of Functions
Related Practice
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