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Problem

Evaluate the integral. $ \displaystyle \int \l…

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Problem 28 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int \sin \sqrt{at}\ dt $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Related Topics

Integration Techniques

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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.

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Watch More Solved Questions in Chapter 7

Problem 1
Problem 2
Problem 3
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
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72
Problem 73
Problem 74
Problem 75
Problem 76
Problem 77
Problem 78
Problem 79
Problem 80
Problem 81
Problem 82
Problem 83
Problem 84

Video Transcript

Let's start off here by doing you use up that here, Do you? Using the chain rule from capitals. We could rewrite this as a over to you DT. And let's go ahead and solve this equation for DT. So we have dt equals to you over a to you and that's whole replace over here in the original integral sign you replaces the first term and then here for DT We can use this so we have a new integral and let's pull out that constant. So on the downside, we do have this extra factor in front of the sign that we didn't have in the original. But the expression inside the sign is now much easier to the OS. So for this in a girl here, you can do integration. My parts Let's go ahead and take w to be you t w equals Do you t v equals sign you do you the equals negative co sign you and then you have to over a recall the formula for integration by parts You've even this case w v minus inner growth e d w. So using that formula here, there's a W times me and then in a girl Z and then DW So this is just negative to you, coz I knew over, eh? Plus two over, eh? Sign you And that was good. And add that sea and finally and the last step here is suggest go to your original substitution and took plug in okay with to plug in U equals radical eighty season. So that's the first term and then for the second term, and that's your final answer.

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Top Calculus 2 / BC Educators
Heather Zimmers

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Harvey Mudd College

Samuel Hannah

University of Nottingham

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
Recommended Videos

04:06

Evaluate the integral. $$ \int \sin \sqrt{a t} d t $$

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