Evaluate the integrals. ∫e^-y cosy d y

Name: Problem 22, Chapter 8: Thomas Calculus
Uploaded: 2020-09-06T20:28:34-08:00
Duration: 3 min 47 s
Channel: Kyler Gray
Description: Evaluate the integrals. ∫e^-y cosy d y

Key Concepts

Integration of Exponential and Trigonometric Functions

Integrals that involve a product of an exponential function and a trigonometric function, such as e^(ay)cos(by) or e^(ay)sin(by), have standard solution methods. One common approach is to use a formula derived from repeated integration by parts, or alternatively, use complex exponential representations (via Euler’s formula) to convert the problem into integrating a single exponential function. This concept is key in solving various differential equations and evaluating integrals in both theoretical and applied mathematics.

Integration by Parts

Integration by parts is a fundamental technique used to integrate the product of two functions by using the formula ?u dv = uv ? ?v du. This method is especially useful when one of the functions becomes simpler upon differentiation and the other is easy to integrate. In problems involving the product of an exponential and a trigonometric function, integration by parts can be applied twice to derive a solvable equation for the original integral.

Solving Recursive Integral Equations

When applying integration by parts repeatedly to integrals involving products such as an exponential function and a trigonometric function, the original integral may reappear on the right-hand side of the equation. This creates a recursive equation which can be solved algebraically to isolate and determine the desired integral. Recognizing and solving these types of recursive relations is an important step when applying integration by parts to these kinds of integrals.

Transcript

00:02 For problem 22, we need to integrate e to the negative y, cosine y, d .y.

00:05 And so we're going to have to do this through integration by parts.

00:08 So let's make u equal to cosine y, cosine y, and du will equal negative sine y, dy.

00:17 And then we're going to make dv equal to e to the negative y, dy, which means when we integrate that, that will just give us negative e to the negative y, and there we go.

00:30 So now we can go ahead and piece this together and we get negative e to the negative y times cosine y minus the integral v times du.

00:41 So negative, we're going to pull out that negative, but then also this is negative.

00:45 So both of those will cancel and we'll still be left with just a negative e to the negative y, sine y, d y.

00:54 And so we now are back to looking like a very similar integral.

00:57 So we're going to have to do this use substitution one more time.

01:00 So let's do a second level of it.

01:03 So u, once again, will equal sine y this time.

01:06 Sign y, du will equal cosine y, d .y.

01:11 We have dv equaling e to the negative y, dy, and v will then equal negative e to the negative y.

01:20 So we can now piece this together for yet a second time.

01:24 So we get negative e to the negative y times cosine of y minus, and then let's distribute this negative.

01:31 So we're going to write all of this inside of it first.

01:33 So u times v, so negative e to negative y, sine y, right? and then minus the integral of v times the u.

01:45 So the negative will come out and make this positive.

01:48 So negative becomes positive, e to the negative y, cosine y, d y...

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