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Evaluate the integral. $ \displaystyle \int_0^…

03:25

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Problem 16 Easy Difficulty

Evaluate the integral.

$ \displaystyle \int_0^{\frac{\sqrt{2}}{2}} \frac{x^2}{\sqrt{1 - x^2}}\ dx $


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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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Top Calculus 2 / BC Educators
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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

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Problem 9
Problem 10
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Problem 16
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Problem 18
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Problem 24
Problem 25
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Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
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Problem 38
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Problem 53
Problem 54
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Problem 83
Problem 84

Video Transcript

Let's use a trick substitution for this integral because the denominator inside the radical is of the form a squared, minus X squared. I would go ahead and use X equal scientific era, then DX. It's because I'm dead so we can rewrite this integral. And we do have a definite on girl here, So we do have the opportunity or not necessary. It's not required, but it can simplify your work to rewrite these limits of integration in terms of fatal if possible. So here plug in X equals zero and then recall when we do it. Troops up of this form signed data you require. That data is in between negative, however too entirely too. So the only solution to this equation and dishonorable up here is state a equal zero. So our new lower limit will still be zero and then plug in the upper limit for X in. The only solution in this interval that makes this possible is PIRA for so that's our new upper limit. And then we see X Square up here in the numerator, so that becomes science squared after the substitution and then we have DX. So that's co sign D data. And then in the denominator, we have one minus X squared inside. The radical selections come to the side to evaluate this and then use the protector in Toronto for sign and co sign to write. This is co sign squared, and then you can evaluate. This is just cosign data. So that's your denominator. And then we could go ahead and cross off those one half. So here we're left with just pie before and then we have just signed squared only. So here I would use tohave angle formula for science. Queer. That's one minus co sign to date over, too. And then here I'm just pulling out the one half and then I have one minus course I into data left over. So that's one half. Now it's in a very that's data minus scientist eight over, too zero pyro for. And then let's just go ahead and plug those and points in so plugging Piper for first. So we have a pilot or for minus sign pi over too, over two, and then when we plug in zero both terms or zero, we have zero minus sign zero over to, so there's nothing to subtract and then sign a pie over to is just one. So we just have pav rate. So we write this pie over eight and then minus one over four, and this is after we did distribute the one half, and that's your final answer.

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Top Calculus 2 / BC Educators
Anna Marie Vagnozzi

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Heather Zimmers

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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.

Join Course
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