7–84. Evaluate the following integrals.
53. ∫ eˣ cot³(eˣ...

Question

7–84. Evaluate the following integrals.

53. ∫ eˣ cot³(eˣ) dx

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate the integral of E2X multiplied by the tangent of E 2X cubed with respect, the X. Now we can evaluate this integral by using substitution. So first, let's let you be equal to E to the 2X. Well, in that case, the derivative of u with respect to X is going to be equal to 22 X, which implies then that DU is going to be equal to 2 E 2 X D X. Or we can even say that half of the u is going to be equal to e 2 X D X, and we wrote it this way because if we compare it to our integral, notice that we have e to the 2XDX in our integral, OK. So now, since we have an expression for you and half of the EU, then we can substitute you into our integra. So by substituting you. We can now rewrite the integral of e to the 2 X and cue to the 2 XD X. Yeah. As the integral of 10 cubed of u multiplied by DU divided by 2 or a half of 1. Cubed of you with respect to you. So now we can integrate with respect to you here. Now for a time cubed of you. Notice that we could rewrite it. OK, so now we're trying to figure out what this integral is and the first step is for us to rewrite it as the tangent of you multiplied by. The square of the tangent of you, OK? And remember that by definition that means then that we could rewrite this as the C can squared of you minus 1. So what this means then is that our integral, if we were to put back this expression into our integral, OK, then what we're really saying is that we're finding a half of the integral here, remember, no, we could, uh, distribute our terms so we're finding the integral of the tangent of you multiplied by the square of the C of you with respect to you. Minus, OK, 1 multiplied by expression here, so that's the integral of the tangent of you with respect to you. So at this point, no, we have two terms, OK, that we are integrating separately. So let's see if we can integrate both terms and then combine our results. No, I'm gonna set up two sides of or. Page here, OK, to integrate both terms. Yeah And I think I'll make the side for our first term a bit longer because we may have to do a bit more, OK? So for our first term, the integral of 1 U 6 squared U with respect to you, again, we could use substitution. So let's let W be equal to the tangent of you. Well, in that case, that means the derivative of W. OK. With respect to you, it is going to be equal to the c 2d of you, OK? And in that case, Then that means that the derivative of you if we're solving for DU, it's going to be equal to. The derivative of or DW divided by 6 squared of you. OK. So now, that means then if we were to substitute this into our integral, then it's going to become the integral of W multiplied by 6 squared U. D U R D W divided by sex squared U. So you can see why I still kept this expression here because sex squared U cancels easily and no, that leaves us with just the integral of W with respect to W. In that case, we know that that integral will be a half of W2d plus C1, a constant. And we know what W is, so we could rewrite this in terms of you as a half of the square and U plus C1. So this is the integral for our first term. Next, for our second term, the integral of the tangent of you, OK. Well, we know that that integral is simply equal to the natural log of the absolute value of the scan of you, OK, plus a constant C2. So now that we've integrated both terms, OK, then we can combine them. So now combining both terms. OK, for combining both terms. Then it's not going to become. A half of the integral of the first term. F times square U plus C1 minus the integral of the second term, which is. The natural log of the C, the absolute value of the C can of U plus C2. Now let's simplify. First we can combine C1 and C2 into some constant C, and if we distribute to our half, then this is going to be a quarter of the square of the tangent of U, OK, minus 1/2 of the natural log of the absolute value of the c kind of you, OK, plus our combined constant c. And now that we are at this point, remember earlier we said U E equals E 2 X. So if we put that back into our expression, this is now going to be a quarter of the square of the tangent of E 2 X. Minus a 1/2 multiplied by the natural log of the absolute value of the scant of e to the 2 x plus C. Thus this is the integral of e2x multiplied by the tangent cubed of e2X with respect to X. Thanks a lot for watching everyone. I hope this video helped.

7–84. Evaluate the following integrals.
53. ∫ eˣ cot³(eˣ) dx

Key Concepts

Integration

Exponential Functions

Trigonometric Functions

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