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Evaluate the integral. $ \displaystyle \int_0^…

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Problem 9 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_2^3 \frac{dx}{(x^2 - 1)^{\frac{3}{2}}} $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

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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Problem 16
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Problem 32
Problem 33
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Video Transcript

Let's compute the integral from two to three of one over exploit menace once of the three have power. So we see in the denominator the thing inside the radical is of the form X squared minus a square. So we should use the tricks up. X equals a second data. And in our problem, since a square is one that means a is one. So we have seek and data. Then the exes seek and their attend data. So at this point observed that we do have a definite integral We have these limits here two and three. So it may be possible to re right those limits in terms of fate of values. But in order to do that, you would have to be able to solve, for example, on X history, equal seek and data, and you have to be able to solve this for data. And this is not a common value that we memorized seemed unit circle. So it is possible to find data. It will require drawing triangles out. So since we have to draw the triangle, we might as well just not find the date of values. Then we could back substitute from X and then we could go back to use our original limits in the final answer. So here also, we have X squared minus one. So this is C can square data minus one to the three house. So tan square data to the three house and that becomes Stan Cube. So this is our new denominator after we simplify it. So since we don't know what those data values are in the limits of integration, let's just call them Seita one and data too. We don't need those for now. Our numerator is the ex seek and data tan data, and we just simplified the denominator a few moments ago. And that's tan. Cute. Okay, so we could go ahead and cross off one of those tangents. Then we're left with ten square in the bottom. So we have seek and data. And then since one over tangent is co sign over sign, we have co science where data over science, where data And then we could use the fact that she can't data is one of her co sign. So we could letting cancel off one of those co signs and we're left over with Cho Santa over sine squared data. So for this General, we could use the U substitution for now. Again, let's just ignore the limits of integration. Let's take for this integral. Here, you two be signed data then do you, as co signed data, even a So we have in a rule, Cassandra, over Science Square is integral of one over you square, do you and use the parole for this one to get negative one over you plus C. So I'm running out of room here. Let's go to the next page. We have we left off negative one over you plus e using our u sub. This's negative one over. Scient, ADA Plus e. And we have those limits data one of data to which we don't want to use because we don't know what they are. So at this point, we should use the triangle. Teoh, right? Everything back in terms of X. So our recall, our original substitution tricks up was X equals seek and data. If you want, you can think of this is X over one. So C can is high pound news divided by Jason so x over one and then used a diary in the room to find age. The remaining side H where plus one is Explorer. So H is the square root of X squared minus one. And then from this triangle we see that signed data is each over X. So X squared minus one squirmed out of this all divided by six. Plug this into our expression. We have negative one over sign and then sense for would have a definite noble There's no need for the sea here. So let's use ignore the sea that will go away It'Ll cancel out when we subtract. Since we're back into our variable X weaken goes for original limits of integration two and three. This is why it wasn't necessary to find Kate on one data too. So it's simplify our our expression. Yeah, Negative. X overexploited minus one. And we're plugging in two and three. So plug in three. First, we have negative three. Oh, our scrutiny minus negative Two. Okay, over squaring off here We'LL have four minus wants is for you three And this could be simplified a little bit. Things come down here so I could rewrite the squared of eight. As to room two, these two minus cistern into a plus. And here it's just a matter of how you want to write the final into. If you're if you don't liketo have radicals in the bottom, you can really congressional eyes that Then I'm under here, and then you could also rationalize the denominator on the first time. So you have negative three times route, too, over four plus to room three. It's a times to me over three, and this is a good place to stop. That's your final answer.

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Top Calculus 2 / BC Educators
Catherine Ross

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University of Michigan - Ann Arbor

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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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