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Evaluate the integral.

$ \displaystyle \int \frac{1}{x^3 \sqrt{x^2 - 1}}\ dx $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Integration Techniques

Campbell University

Oregon State University

Harvey Mudd College

Idaho State University

Lectures

01:53

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.

27:53

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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Evaluate the integral.…

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Evaluate the indefinite in…

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Evaluate the definite inte…

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02:22

Evaluate the integrals…

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04:51

Find the Integral of \int …

to evaluate this inner girl by trying to trick substitution X equals seeking data and the motivation for this This choice of a treat so is based on not necessarily the radical, but more so the term X squared minus one. So this is a number of the form X squared minus a square. And when you're in this case and you want to use shrinks of, then this is the tricks of that you should use. So here, differentiate. So that's already inside. And then now we can rewrite this integral So dj X up top. I can replace that with this. So I have seeking data tentative data up top and then on the bottom C can cube, and then we'LL have square root seek and square data minus one. Now, here we know that seek and square data minus one equals stance. Where there are this's one year Pythagorean identities. So when we take the square room, we just get tan data and so we can go ahead and cancel this tan data here with this tan date up there and then we could cross off this. He can't with one of these and we have two left on the bottom. So we have one over c can squared and we could rewrite that as the integral of co signed square. Really? No. And then used a half angle identity for co sign. And we could integrate both of these terms If this through data is bothering you feel free to do use up here. You can do u equals to data and we have one half data and then we have sign to theatre and we'll have a to from this over here and then we'LL get another two from the use of And then here we can rewrite this using the double angle formula for sign. And then we could cross off that too, with one of the other two's. So we have one half data. So let me just write that a state over to and then signed Data Co signed data over two plus c. Thanks. Now let's evaluate each of these three terms, and these will all come from our use of So, for example, if I take in verse, he can on both sides I find data so I can rewrite this answer as C can't members of X over to you urgent. And then now I need to evaluate, sign and co sign. So this is where I go to the triangle. So we did seek and data equals X over one. So that gives us X divided by one. If recalling this angle over here data no. And then by Pythagorean dirham, we get the remaining side. So now we could go ahead and find sign Ako side. So, science officer, over half pound news. So we have over X for sign and then for co sign Jason over. Heaven knows it's one over X and then we still have our two down there and the last appears Just simplify this answer a little bit. She can inverse effects over too square root, X squared minus one over to X squared. Plus he and that's our answer.

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