The integrals in Exercises 1–34 converge. Evaluate the integra...

Question

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫₁^∞ dx / [x√(x² − 1)]

Accepted Answer

Hello. In this video, we are going to be evaluating the following improper integral. The integral given to us is the integral from 1 to infinity of dx divided by x2 multiplied by the square root of x2 minus 1. Now, the first thing we want to do is you want to go ahead and make a substitution for the infinite limit. By using the rules of the improper integral, we can use a variable to replace the limit. Let's say we have the variable b. Then we can create a limit as B approaches infinity of the indefinite integral 1 tob of 1 divided by x2 multiplied by the square root of x2 minus 1 d x. So, ignoring the limit for now, how can we approach this improper integral? Well, because we have a square root in the denominator, this is an indication that we should approach this using trigonometric substitution. And if we were to use trigonometric substitution, this will be of the form x is equal to a sequent of theta. Here a is going to equal to 1. And our X term is just going to be X. So we will have the substitution x is equal to sin of theta. And that means that by taking the derivative of the statement dx is going to equal to sequent theta tangent theta. The theta. So let's go ahead and apply these substitutions for now. So if we go ahead and apply the substitutions and ignore the bounds for now, we will have. The integral Of seeking theta. Tangent theta. Divided by sequent square theta. Multiplied by the square root. Of sequin squad theta minus 1 d theta. Now let's go ahead and simplify. First, we see that we have a multiple of sequin the in the numerator and 2 multiples of it in the denominator, so we can eliminate 1 sequin theta from both the numerator and denominator. Now sequin 2 minus 1 is equivalent to tangent 2 theta. And the square root of tangent squared theta is equivalent to just tangent of theta. So this entire quantity is going to simplify out with the tangent theta in the numerator. So, if we go back to the original integral, we will have the limit as B approaches infinity of the definite integral from 1 to B. Of 1 divided by sequent theta d theta. Now here, sin theta is equivalent to cosine of theta. So by using our Pythagorean identity, we will have the limit as B approaches infinity from 1 to B of cosine of theta d theta. And by taking the integral, the antiderivative of this is going to be the limit as B approaches infinity. Of sin of theta Integrated from 1 to B. Now here, what we need to do is we need to resubstitute sin of theta back into terms of X, and we can do this by using a right triangle and our original substitution for the trigonometric substitution. So again, recall that we said x equal to sequent of theta. By rewriting the statement, we can set sequin of theta equal to x divided by 1. And recall that the identity for seeking of theta is the hypotenuse divided by the adjacent side with respect to theta. So if we were to create a right triangle. And to set our angle of theta. Here then Sein of theta tells us that we have x for the value of the hypotenuse and 1 for the value of the adjacent side to theta. Now we can solve for the missing side of this right triangle by using the Pythagorean theorem. That will give us X2 is equal to 12, which is just 1, and we will call the missing side B, so we will have 1 + B2. Solving for b will give us b is equal to the square root of x2 minus 1. So that's going to be the missing side of the right triangle. But why does this matter? Well, this matters because now if we solve for sine theta of this right triangle, we can replace sine theta with the opposite over hypotenuse. So that is going to give us the limit as b approaches infinity. Of the square root of x2 minus 1 divided by x, and this is going to be evaluated from 1 to b. So now if we evaluate this at the boundaries, we will have the following equate following expression. We will have the limit as b approaches infinity of the square root of b2 minus 1. Divided by b minus the square root of 1 squared, which is just 1 minus 1 divided by 1. Here the second term is going to simplify to be zero, and if we are to apply the infinite limit to the first term, the first term is going to simplify to be 1. So the value of the definite integral to the original problem is just going to be 1, and that is going to be the solution to the problem with the overall solution being option B. So I hope this video helps you in understanding how to approach improper integrals, and we will go ahead and see you all in the next video.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₁^∞ dx / [x√(x² − 1)]

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The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₁^∞ dx / [x√(x² − 1)]