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Evaluate the integral.

$ \displaystyle \int e^{2 \theta} \sin 3 \theta d \theta $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Missouri State University

Campbell University

Harvey Mudd College

University of Nottingham

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

05:12

01:45

Evaluate the integral.…

06:20

03:42

02:01

Evaluate the integrals.

03:59

07:21

03:16

06:54

Evaluate the following int…

05:51

The problem is you evaluated the integral into 2 theta times sine 3 theta d theta for this problem we will use the method of integration by parts, so the formula is integral: u times, prime dx is equal to? U times, v minus the integral of? U. Prime times v d x for our problem, we can let? U is equal to 22 theta and the way prime is equal to sine 3 theta. Then prime is equal to 2 times e to 2. Theta and v is equal to 1. Third cos negative 1. Third. Cosine 3. The now this integral is equal to u times v, but this is negative. 1. Third to 2 theta sine 3 theta minus the integral of prim times v. So this is negative. 2. Third, this plus 2 third times to 2 theta cosine 3 theta d. Theta point now for this integral we can use also use the method of integration by parts that the? U is equal to e to 2 theta nopolis equal to cosine, 3 theta? Then? U prais equal to 2 times e to 2 theta and v is equal to 1 third times sine 3 theta. Now this is equal to negative 1 over 3 into 2 theta cosine 3 theta plus 2 over 3 times. U times so. This is 1. Third to 2 theta sine 3 theta minus integral f you prime times, so this is 2 third to 2 theta times sine 3 thetasothis is equal to negative 1 third into 2 theta cosine 3 theta plus 2 over 9 into 2 theta am sine 3 theta minus 4, over 9 integrite 2 theta in sin, 3, theta d, theta now locatintegral. This is the same as this 1, so we move this term to left hand side and we have 1 plus 4. Over 9 is integral into 2 theta times sine 3 theta d. This is equal to negative 1. Third, third into 2 third 2 times theta cosine 3 times theta plus 2 over 9 into 2 times theta times sine 3 times thetaso this integral this constant is 13 over 9 point. So this integral this integral to 2 theta sine theta d theta is equal to negative 1 third times on, so we multiply 9 over 13 for both sides, and this is negative. 3. Over 13 e to 2 theta cosine, 3 theta plus 2 over 13 into 2 theta sine 3 theta and plus to const.

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