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Evaluate the integral. $ \displaystyle \int_0^…

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Problem 8 Easy Difficulty

Evaluate the integral.

$ \displaystyle \int_0^{2\pi} \sin^2 \left(\frac{1}{3} \theta \right) d \theta $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

Related Topics

Integration Techniques

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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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answer the following?

Watch More Solved Questions in Chapter 7

Problem 1
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Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
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
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
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Problem 45
Problem 46
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Problem 48
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Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
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Problem 70

Video Transcript

this problem is from Chapter seven, Section two. Problem number eight in the book Calculus Early. Transcendental lt's a Definition by James Store. Here we have a definite overall from zero to pi. Sign square of data over three with respect that data. So for this problem, since we haven't even power on the sign and there's no co signs present, it's going to be best to use this trick in a metric identity, which says Sign square of X is one minus coastline of two x divided by two. So whatever the input is on, the sign over here in co sign multiply the same thing by two. So let's use this time to rewrite this. Integral as the integral from Sierra two pyre, they have a one half minus co sign of two times one third data. So we have ah to that over three. Also, divide this turn by, too, because we're dividing by two over here and now we can evaluate you should these integral separately. So the interval of one half with respect the data simply they divided by two. And for this integral over here we can apply you, so let's take you to be to date over three. So using this use of been evaluating this underworld the second integral we'LL tend sign of to date over three times three over too. And we're evaluating thes this expression at zero in super. So let's plug into my first for data thirteen two pi divided by two minus sign of four pi over three. And then here Let's simplify this fractions. We have three over two and then we have another two on the bottom. So let's just multiply this by three or four minus. Now we plug in Seo for data. So having zero over too, minus sign of zero over two times three over. Horrible. Did you write that? So here we have zero and from trigonometry with no sign of zero zero. So here the twos, cancel. We just have a pyre. The level minus and in sign of four pi over three is negative. Rule three and divide that by two. And then we still have three over four. So here we could just multiply these denominators together to get eight on the bottom and cancel out this double negative tub tin PIDE plus three, Route three divided by. And that's your final answer

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Calculus: Early Transcendentals

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Related Topics

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Top Calculus 2 / BC Educators
Heather Zimmers

Oregon State University

Kayleah Tsai

Harvey Mudd College

Michael Jacobsen

Idaho State University

Joseph Lentino

Boston College

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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answer the following?

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