Evaluate the integrals in Exercises 67–74 in terms of
a....

Question

Evaluate the integrals in Exercises 67–74 in terms of

a. inverse hyperbolic functions.

71. ∫(from 1/5 to 3/13)dx/(x√(1-16x²))

Accepted Answer

Welcome back everyone. Evaluate the following definite integral. Express your answer in terms of an inverse hyperbolic function. Integral from 1/7 to 1/6 of d x divided by x square root of 1 minus 25 x 2. For this problem, let's remember that the integral of d u. Divided by you. The square root of a 2 minus u2. Is equal to -1 divided by a. Inverse of sesh of u divided by a plus a constant of integration c. So in this case, to achieve this type of form, we're going to introduce you substitution. Specifically, let's understand that 25 x squared is equal to 5 x 2 using the properties of exponents. So we're going to set u equals. 5 acts. Now let's find the derivative of you with respect to x. That's the derivative of 5 x, which is 5. We can now show that dx is equal to duu divided by 5 or 1/5 du. The next step is to change the limits of integration, so the lower limit u1 is going to be 5 multiplied by 1/7. Which is 5 divided by 7. The upper limit of integration is going to be U2, which is 5 multiplied by 1/6, and that's 5 divided by 6. So we can now go ahead and rewrite the integral in terms of you and DU. We're integrating from 5 divided by 7 up to 5 divided by 6. DX becomes 1/5 DU. In the denominator we have x, which becomes u divided by 5 because u equals 5 x, so that x is equal to u divided by 5, right? Followed by a square root of 1 minus. 25 x squared is basically U squared. Now, let's simplify. 1/5 divided by 1/5 can be basically canceled out, right? Because we get one. And this gives us. The integral from 5 divided by 7 up to 5 divided by 6 of the. Divided by you. square root of 1 minus u squad and we already know its value. It is -1 divided by a and a in this case is 1 because 1 squared. Corresponds to a squared. So we get 1 divided by 1, which is -1. In verse of sash. Of you divided by A, so that's you divided by 1 or basically you. So our integral is negative inverse of of you from 5 divided by 7 up to 5 divided by 6. And now we can evaluate the result. So that'd be negative inverse of se of. 5 divided by 6 minus negative inverse of sesh of 5 divided by 7 and now we can simplify, right? Distributing the negative sign we get. Positive inverse of sesh. Of 5 divided by 7. Minus In verse of Sesh. Of 5 divided by 6. That's our final answer, and thank you for watching.

Evaluate the integrals in Exercises 67–74 in terms ofa. inverse hyperbolic functions.71. ∫(from 1/5 to 3/13)dx/(x√(1-16x²))

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Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
71. ∫(from 1/5 to 3/13)dx/(x√(1-16x²))