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


∫₀^{π} 2^{sin x} · cos x dx

Verified step by step guidance

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 \(\cos x\).
2. Use substitution to simplify the integral. Let \(u = \sin x\). Then, compute the differential \(du = \cos x \, dx\).
3. Rewrite the integral in terms of \(u\): since \(du = \cos x \, dx\), the integral becomes \(\int_{u=\sin 0}^{u=\sin \pi} 2^u \, du\).
4. Evaluate the new limits of integration: \(\sin 0 = 0\) and \(\sin \pi = 0\), so the integral becomes \(\int_0^0 2^u \, du\).
5. Notice that the limits of integration are the same, which implies the value of the integral is zero without further calculation.

Key Concepts

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.

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.

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.