Verify the integration formulas in Exercises 37–40.
37. ...

Question

Verify the integration formulas in Exercises 37–40.

37. a. ∫sech(x)dx = tan⁻¹(sinh x) + C

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the integral, the integral of hyperbolic C count to the power of 3 of x dx. OK, so it appears for this particular problem we're simply asked to evaluate this specific integral. So now that we know that we're evaluating this integral. Our first step that we need to take, so this will be step one, is we need to recall and use the following identity that states that hyperbolic sec to the power of 3 of x will be equal to hyperbolic secant. Of x multiplied by hyperbolic c can square of x. And we will also need to recall and use our substitution method stating that U for this particular prompt will be equal to hyperbolic tangent of X, and we need to note that DU will be equal to hyperbolic c square of XDX. Thus, we can go ahead and write that I or integral i will be equal to the integral of. Hyperbolic c of x multiplied by hyperbolic c square of x dx, which will be equal to the integral of hyperbolic secant of xDu. Our second step now is we need to express hyperbolic se of X in terms. Of simply, so once again, we're trying to solve. Using our substitution, so we're gonna be simplifying and, and we need to note. That sense, once again, we need to note that sense. Hyperbolic tangent squared of x plus Hyperbolic sea Kent. Of, you know, so once again, hyperbolic c can square of X is equal to 1, and we need to note that since hyperbolicc of X is greater than 0, and this is for real values of X, that means we can simply go ahead and write. So, I mean earlier, I need to be specific. We're expressing hyperbolic se of X in terms of you, and we need to note the following identities, and we also need to note that hyperbolic se of X is greater than 0 for real values of X. And keeping that in mind, we can simply write that hyperbolic se of X will be equal to the square root of 1 minus. Hyperbolic tangent. squad of X, which will be equal to the square root of 1 minus u to the power of 2. So that will mean that I is simply equal to the integral of the square root of 1 minus u to the 2DU. And for our third step, this is step 3, we need to compute the integral of the square root of 1 minus u to the power of 2 DU, and one convenient method to do that, to do this, is to use the substitution method using specifically U is equal to sin theta, noting that DU will be equal to cosine of theta. The theta. And this will give us the integral of the square root of 1 minus u to the power of 2DU, which will be equal to the integral of cosine squared of theta. D theta, which will be equal to 1/2 multiplied by the integral of parentheses of 1 plus cosine of 2 theta d theta. Fantastic. And this will be equal to simply 1/2 multiplied by parentheses of theta plus 1/2 multiplied by sin of 2 theta plus c or c denotes some constant variable. So, once again, we now need to return to our u variable, and we need to note that theta will be equal to inverse sin of U, and we need to note that. Sine of 2 theta will be equal to 2 multiplied by sine of theta multiplied by cosine of theta, which will be equal to 2 multiplied by u multiplied by the square root of 1 minus u to the power of 2. And this will yield the standard antiderivative. So we can go ahead and write the integral of the square root of 1 minus u to the power of 2, DU, which will be equal to 1/2 multiplied by parentheses of u multiplied by the square root of 1 minus u to the power of 2, plus inverse sin of u. Plus C And for our 4th and final step we need to substitute back in u is equal to hyperbolic tangent of x, and we will do this by recalling and using that the square root of 1 minus hyperbolic tangent squared of x will be simply equal to hyperbolic c can of x. So that will mean that i will be equal to 1/2 multiplied by parentheses of hyperbolic tangent of x multiplied by hyperbolic sec of x plus inverse sin of. Hyperbolic tangent of X plus C, and that's it. That is our final answer. Hooray, we did it. So going back to look at the prom itself to recap what we've learned, we could safely say that our final answer without a doubt will be once again. Drum roll. 1/2, so it's gonna be 1/22. Multiplied by parentheses of hyperbolic tangent of x multiplied by hyperbolic c x plus inverse sign of hyperbolic tangent of x plus c, noting once again that c denotes some constant variable, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Verify the integration formulas in Exercises 37–40.37. a. ∫sech(x)dx = tan⁻¹(sinh x) + C

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Verify the integration formulas in Exercises 37–40.
37. a. ∫sech(x)dx = tan⁻¹(sinh x) + C