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Find the area of the region under the given curve…

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Problem 63 Hard Difficulty

Use the substitution in Exercise 59 to transform the integrand into a rational function of t and then evaluate the integral.

$ \displaystyle \int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 + \cos x}\ dx $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Related Topics

Integration Techniques

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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

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Problem 9
Problem 10
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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 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

Video Transcript

Let's use the substitution from exercise 59. That's this equation over here. And these were the results of that exercise. We ended approving these equations here based on the substitution. So using that substitution, let's go ahead and rewrite this in overall. So first I see that I can rewrite this as to sign times Go sign. So let me just go ahead and pull out that too, and then let me instead of writing sign, go ahead and use our right hand side over here. So that's two t atop So that sign and then for co sign we have another equation here. So that's our numerator. We have two times sine times co sign and we make that integral longer for t X. It's just this over here. So this is all in our numerator and then oops, there should be a d. T. At the end here too. Uh huh. And then on the bottom two plus co sign. So two plus one minus t squared over one plus t squared. And because we're in a new variable teeth, we should go ahead and change those limits of integration. So how to do that? Just plug in these original limits into X in this equation over here and you'll get your T limits so the lower limit will be t equals tangent of 0/2. So it's J zero for the upper limit. And then if we do pi over two over two, that's power for that's one. So there's an upper limit. Let's go ahead and simplify this. Multiply out all these tools up here. Get an eight. Pull that out. So this is our numerator. After we multiply all those quadratic together, we have a third power. That's the numerator the denominator. Go ahead and add those fractions together on bottom combine like terms. So this is our denominator and then go ahead and cancel. So here we could cancel this with one of these. That's where the two is coming from. Over here. Now I'm running out of room here. Let me go to the next page. So we have eight and then go ahead here and let's right. This is yeah. So this is just for the partial fraction decomposition for the previous rational function that we just had on page one. Yeah, So now there's three inaugurals for the first one here. Just go ahead and do a U sub and then for the second, very similar. And what should we do for the last one? Same as the original one. Mm. Okay, so here's the first integral eight. But then we also have that, uh, minus. Here. Let's pull out that minus. And we have a two here to divide by after the use of you could change those limits of one and two for the next inaugural. Eight in the two again, this time from 3 to 41 over you, do you? And then for the last 1 8/2, 1 to 2. Do you ever u squared and we know how to evaluate. All these here is negative for l N 1 to 2 for Ellen three and four. And then we have minus four over you from 1 to 2. Let's go ahead to the next page and plug those in. So we have negative for l n two plus four Ln four minus four l n three and then minus two plus four. And this could finally be written as two plus four Ln for over six, and then maybe go ahead and just cancel. Simplify that fraction, and that's our final answer.

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Related Topics

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Top Calculus 2 / BC Educators
Catherine Ross

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Anna Marie Vagnozzi

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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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