In Exercises 27–40, use a substitution to change the integral ...

Question

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.

∫ dt / (tan(t)√4 - sin^2(t))

Accepted Answer

Welcome back everyone. In this problem we want to use a table of integrals to evaluate the integral of dx divided by cut x multiplied by the square root of 9 minus cost square x. Now before we do that, let's first rewrite cut x in our integrand. Recall, OK, that the reciprocal of cut x is really the tangent of X. So with that in mind, then we could rewrite dx, the integral of dx divided by cut x multiplied by the square root of 9 minus cut square x. As the integral of 10 x divided by the square root of 9 minus square x with respect to x. Now before we go for our table of integrals, let's first try some substitution to simplify our integral. Now by substitution. OK Let's set you. So the cosine of x. Then that means the derivative of u with respect to x would be negative sin x, so this means duu would be equal to negative sin x dx. And with that That means the tangent of X as well by the definition of the tangent. Is going to be the sin of x divided by the cosine of x. OK, so sin x divided by cos x multiplied by dx. If we rewrite this using the fact that u equals x and duu equals negative sin x dx, then this would be equal to duu divided by u. So now we have 10 xDX in terms of you and we have cost squared x in terms of you or we have cost x in terms of you. So with that in mind. Then that means our integral. Of 10 x divided by the square root of 9 minus cost squared x with respect to x can be rewritten as the negative value of the integral of d u divided by u multiplied by the square root of 9 minus u 2. Now, let's take a break, let's take a quick stop here. Does this look like anything we've ever seen before? Well, if we use the standard table for integrals, OK, we find, OK, that, well, let me write it below here instead. We find that it looks really similar to this integral, the integral of duu divided by u multiplied by the square root of a 2 minus u2, where a 2 would be 9. And that integral is equal to 1 divided by a multiplied by the natural log of a minus the square root of a 2 minus u2 all divided by u, and that's the absolute value for our natural log that is plus c. Now in this case a squared as we said would be equal to 9, which means that a would be equal to 3. So if we use this idea, OK, thus that means we could rewrite the negative value of d u divided by u multiplied by the square root of 9 minus u 2 as 1/3. Of the natural log of the absolute value of 3 minus the square root of 9 minus u 2 divided by u. Plus C Now we can simplify to rewrite this in terms of x. Remember earlier we said that u equals cost x, OK, so we could rewrite this then as -1/3 of the natural log of 3 minus the square root of 9 minus cost square x. Divided by class X, OK, plus C. Now remember here that when we're talking about logs, let's write this property on our right, recall, OK, that if we have a LNB, we could rewrite it as LNB to the power of A. OK, that's a property of logs. So in this case, since we have -1 as a coefficient of ln, we could rewrite it in our as our power. It's argument raised to that power. In other words, we can say here. That this is going to be 1/3 of the natural log of the absolute value, well, let me put this in a bracket in a parentheses, the absolute value of 3 minus the square root of 9 minus k squared x. Divided by cost x, OK, and this is all raised to the power of -1. Well, if something is raised to the power of -1, really we're saying it's the reciprocal of the term. So for the reciprocal that means then. That we could rewrite this as 1/3 of ln of the absolute value of cost x divided by 3 minus the square root of 9 minus cost squared x and we have all of this plus some constant c. So basically we've simplified our logarithm to obtain an equivalent but slightly simpler form. Therefore, this tells us, let me put this in back here, that the integral of d x divided by cut x multiplied by the square root of 9 minus cost squared x is equal to 1/3 of the natural log of the absolute value of cost x divided by 3 minus the square root of 9 minus cost squared x plus c. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.∫ dt / (tan(t)√4 - sin^2(t))

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In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.
∫ dt / (tan(t)√4 - sin^2(t))