Evaluate the integral. ∫e^-θ cos2 θd θ

Name: Problem 14, Chapter 6: Essential Calculus Early Transcendentals
Uploaded: 2020-05-15T14:09:03-08:00
Duration: 6 min 20 s
Channel: Robert Daugherty
Description: Evaluate the integral. ∫e^-θ cos2 θd θ

Key Concepts

Integration Techniques for Exponential-Trigonometric Functions

This concept involves evaluating integrals where both an exponential function and a trigonometric function are present. A standard approach uses a known formula derived from repeated integration by parts or through the complex exponential representation. Such formulas typically reduce these integrals to expressions that combine exponential and trigonometric functions, making the evaluation systematic.

Integration by Parts

Integration by parts is a key technique in calculus for integrating products of functions. It is based on the formula ?u dv = uv - ?v du, which allows one to differentiate one part of the product and integrate the other. This approach is especially useful when dealing with products like an exponential function combined with a trigonometric function, as repeated application can lead to relations that simplify or even solve the integral.

Complex Exponential Representation

This concept uses Euler's formula (e^(i?) = cos? + i sin?) to express trigonometric functions as parts of complex exponentials. By converting cosine and sine functions into the real and imaginary parts of an exponential function with an imaginary exponent, one can often streamline the integration process by handling a single exponential function instead of separate trigonometric terms.

Transcript

00:02 6 .1 from 14, the integral of e to the minus theta, cosine of 2 theta, d theta.

00:08 This is a classic problem for integration by parts, an exponential and a cosine or sign.

00:14 They all pretty much follow the same path that you're going to need to integrate by parts twice to get to a solution.

00:20 So we start out, let u equal e to the minus theta, then du is going to be minus e to the minus theta d theta.

00:32 And then dv is going to be the cosine of 2 theta d theta you integrate the cosine to get the sign function so this is going to be one -half the sign of 2 -theta so i can rewrite this integral e to the minus theta cosine of 2 -d -theta is equal to uv so that's one -half e to the minus theta and sine of 2 theta minus the integral of v -d -u so that's going to be plus 1 -half and then the integral of e to the minus theta sine of 2 -theta d -theta and now what i need to do is work on this integral integration by parts so same substitution u is equal to e to the minus theta d -u is equal to minus e to the minus theta d -d -d -v is equal to the sign of 2 theta d theta.

01:55 If you integrate the sign, you get minus cosine.

01:59 So this is going to be minus 1 half cosine 2 theta.

02:07 So let's see what we've transformed into now.

02:11 So i've got the integral e to the minus theta, cosine of 2 theta d theta is equal to 1 .5.

02:20 E to the minus theta, sine 2 theta, plus 1 half, and now it's uv minus vdu, and so this becomes minus 1⁄2, e to the minus theta, cosine 2 theta, and then minus the integral of vdu, so that becomes minus 1⁄2 of e to the minus theta, cosine 2 theta d theta...

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