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Evaluate$$ \displaystyle \int_2^\infty \frac{1}{x \sqrt{x^2 - 4}}\ dx $$by the same method as in Exercise 55.

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$\frac{\pi}{4}$

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Wen Zheng

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 8

Improper Integrals

Integration Techniques

Oregon State University

Baylor University

Boston College

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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Were given an integral. Were asked to evaluate this integral by the same method as in Exercise 55. The integral is from 2 to Infinity. Mhm. Of 1 over x times square root of X squared minus four D X. Okay, What makes this integral special is that it is clearly an indefinite integral. Not only because we have an infinite into grand, That is the limit is improper But also because our function on the inside is discontinuous At x equals 2. So there's really two Points of ah indefinite nous however, We can evaluate this integral by writing it In two different methods. So first We'll use AU substitution. Now we recognize that in the denominator here we have a X squared minus four And we have a square root of that. This suggests maybe some kind of trig substitution. For example, consider X equals Square to 4 or 2 second of t. Then it follows That DX is equal to Find the differential. This is to seek anti tangent. T DT Yeah. And so the integral of one over X times the square root of x squared minus four T. X. Well this becomes The integral of 1 over And then X we know is two times the sequence of T Times The Square Root of This is 4 times Seek and Squared of T -4. And then DX becomes to seek and T tangent T. D. T. At this point. The 2nd tease, cancel out. We have the integral of tangent E over. And then the denominator the 2 is also cancelled out. We have the squared of four and second squared t minus one. This is the same as tangent squared thi this is really two times the square root of tangent squared T. D. T. Using trigonometry. And then you can simplify By taking the square root and we get the integral of Tangent of T over 2 times the absolute value of tangent of T. D. T. Now the question is can we or can we not drop the absolute value? Well, This kind of depends on what domain you're using. So, You know that X lies between 2 and infinity. We also know that X is equal to to seek and T. So second t is going to lie between 1 and infinity. What does this tell us about? T? Well, this is the same thing as co sign. T has to lie between zero and one. This of course means that T. is going to have to lie between negative Pi over 2 and positive pi over 2. And we also have the T. cannot be equal to 0. In fact, can be equal to any multiple of pi. Now, if T lies between negative pi over two and positive pi over two, Well, it's not really clear exactly what the sign of Tangent E is going to be. In fact, since we're just looking at the inverse co sign, We're really only concerned about T between 0 and Pi over 2. For these values of T, It follows the tangent of T Is going to be greater than 0. Therefore are integral simply becomes the integral of one half DT Which is 1 half tea. Now we have it X is equal to two seconds. T so 1/2 of thi this is going to be equal to mhm 1/2 of the inverse 2 of one half X. This is 1 way of looking at it. Um So Now we can split the integral At any number between To infinity. So for example, we could split it at 3. So they are integral from 2 to infinity of one over X times the square root of x squared minus four dx is equal to the integral from 2 to 3 of one over X times the square root of x squared minus four D x Plus. The integral from 3 to Infinity of one over X times the square root of x squared minus four dx. Notice that these are both improper. This is this one on the right is improper because the limit is infinity While the one on the left is improper because the function is discontinuous at X equals two. So to evaluate, will write these as The limit as a approaches to from the right of the integral, Which we know. The anti derivative is now 1/2 in verse second of 1 half of X, evaluated from X equals a 23 plus. And this is not the limit. As B approaches positive infinity of again are anti derivative 1/2 inverse seconds Of 1/2 x From x equals 3 to be So plugging in. This is the limit as a approaches 2 from the right of one half in verse. 2nd of This is 3/2ves minus one half universe. Second of a over to plus The limit. As the approaches infinity of one half inverse second A few over to minus one half inverse second of three halves. So it's easy to see that the one half inverse seeking of three halves these cancel out And so we're simply left with. Well, if we let the approach infinity the universe second be approaches infinity. Well, this is 1/2 times the inverse sequence of infinity minus one half times The Inverse 2nd of one. Now the Inverse 2nd of 1. This is pretty easy. This is yeah. See The same as the inverse co sign of 1, which is zero, So we're simply left with 1/2 times the inverse second of infinity. Now second is going to be equal to infinity when co sign is equal to 0. This is the same as one half times pi over 2 by inverse functions, and this is pi over four. So we get our answer pi over 4.

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