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Evaluate the integral. $ \displaystyle \int e^…

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Problem 17 Medium Difficulty

Evaluate the integral.

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


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Related Topics

Integration Techniques

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Top Calculus 2 / BC Educators
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Video Thumbnail

01:53

Integration Techniques - Intro

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.

Video Thumbnail

27:53

Basic Techniques

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

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Watch More Solved Questions in Chapter 7

Problem 1
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Problem 5
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Problem 7
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Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
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Problem 39
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Problem 74

Video Transcript

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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Related Topics

Integration Techniques

Top Calculus 2 / BC Educators
Catherine Ross

Missouri State University

Anna Marie Vagnozzi

Campbell University

Kayleah Tsai

Harvey Mudd College

Samuel Hannah

University of Nottingham

Calculus 2 / BC Courses

Lectures

Video Thumbnail

01:53

Integration Techniques - Intro

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.

Video Thumbnail

27:53

Basic Techniques

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

Join Course
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Evaluate the integral. $$\int \sin 3 \theta \cos 2 \theta d \theta$$

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