7-56. Trigonometric substitutions Evaluate the following integ...

Question

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

33. ∫ √(x² - 9)/x dx, x > 3

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate the integral using trigonometric substitution of the the square root of X2 minus 49, all divided by X with respect to X, where X is greater than 7. Now for us to evaluate using trigonometric substitution, let's first do a few substitutions here. First, let's let i be equal to the integral of the square root of X2 minus 49 divided by X with respect to X. That will make it easier to refer to our integral going forward. Next, let's let x be equal to 7 multiplied by the sant of theta. That means then that the derivative of X with respect to theater will be equal to 7 theta theta, which means that D X is going to be equal to 7 sec theta and theater theater. Now, we've been able to express using trigonometric substitution, DX, X and or integral. So we have all we need to use trigonometric substitution. So now by substituting. OK. Then this means that our integral. It's going to be equal to the integral of the square root of X2, so that's 76 theta squared. Minus 49 divided by x 76 theta. And now instead of DX we're going to write 7 sec theater, 10 theater, the theater. Now if we think about it here, 7 sec theta cancers. OK, and if we expand our square term underneath the square root. Then we're going to have the integral of the square root of 49 6 square theta minus 49, OK, multiplied by 10 the D theta. Now, if we, uh, if we, if we factorize under our square root, that's gonna give us the integral of the square root of 49 multiplied by 6 square theta minus 1, OK. And the The And now, using what we know about trigonometric identities, recall, OK, that Times Square Theater. Let me write this properly here. 10 square theta is equal to 62 minus 1, OK? So we can use that identity to substitute in our problem because no, when we do that, then we'll end up with the integral of the square root of 49 multiplied by 1 squared the tan the D theta. Now when we simplify that, OK, we know that the square of 49 is 7, so we could rewrite this as the integral of 7 turn theater. OK. The square root of 0 squared theater is turn theta multiplied by turn theta the theater. And 10 theater multiplied by 10 theater equals 10 square theater, so we could rewrite this as 7 multiplied by the integral of 10 squad theater the theater. Which is 7 multiplied by the integral of 62 theta minus 1 with respect to theater, and we wrote it like that because the integral of 6 squared the minus 1 with respect to theater will be equal to 10 theater minus theater, OK? So our integral is 7 multiplied by 10 the minus theta plus C. I know if we expand that, that is gonna be equal to 7 times the minus 7 theta plus C. Now, where do we go from here? Well, no, we can replace all our terms with respect to theta with our X terms. Remember, we said that X equals 7 theta. So now if we solve for theta, that means se theta, OK, is going to be equal to 7th of X, which means that theta is going to be equal to the arcs. OK. The arcC can. Of x divided by 7. Now on the other hand, OK. Remember, we know that ran theta equals the square root, OK. The square root of 6 square theta. -1 OK. So now, since we know what uh X is, or we know what 6 square theta is. And we know what sec theta is, we can substitute it into our problem. So no, that's going to be 7 of X2 minus 1. Now if we start to simplify here, this is going to be equal to. The square root of X2 divided by 49 minus 1. OK. Let me make some more space again on the right because no, this is then going to be equal. To the square root of X2 minus 49 divided by 7. So now we can use these to replace the terms in our final integral. Which means that i is going to be equal to 7 the, so that 7 multiplied by the square root of X2 minus 49 divided by 7. OK, minus 7 the, so that's 7 multiplied by the arc of x divided by 7. Plus C I know when we simplify we know 7 sorry, 77 in our first term, so that's gonna give us the square root of X2 minus 49, OK, minus 7 multiplied by the arc of X divided by 7 plus C. This would be our integral after we've used trigonometric substitution. Thanks a lot for watching everyone. I hope this video helped.

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.33. ∫ √(x² - 9)/x dx, x > 3

Watch next

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
33. ∫ √(x² - 9)/x dx, x > 3