Evaluate the integrals in Exercises 33–52.
∫ cot⁶(2x) dx...

Question

Evaluate the integrals in Exercises 33–52.

∫ cot⁶(2x) dx

Accepted Answer

Welcome back everyone. In this problem, we want to evaluate the integral of cut 4 x 4th with respect to x. Now we can evaluate this integral using use substitution, OK, so by substitution. Let's start by setting u to be equal to 4 x. Then that means the derivative of u with respect to x is going to be equal to 4, OK, which means that we can rewrite dx as 1/4 of duu. OK, so in this problem we've been able to rewrite you in terms of x and DU in terms of dx. So now we can go ahead and put it into our integral. Because this means then that the integral of cut 4 x 4th. OK. Cut for x always to the 4th with respect to x can be rewritten as 1/4 of the integral of cut u to the 4th with respect to u. Now, let's rewrite the integrand in terms of Kose you and cut you using the identity for the square of cut you because recall, OK, recall. That the square of cutu is equal to the square of coseq minus 1. So in that case then a quarter of the integral of cut u 4th with respect to u could be rewritten as a quarter of the integral of cuts u 2 multiplied by cut u square with respect to you. OK. Which, if we expand that a bit more, that will be 1/4 of the integral of cut U2 multiplied by Cose U2. OK, Cose U 2 minus 1 with respect to you. Which is going to be equal to a quarter multiplied by the integral of cut square u coex 2 u. Minus God squared you all with respect to you. Now because we have it this way, then we can split our integral. So really then we could rewrite this as a quarter of the integral of cut squared u. OK, squareu multiplied by cosex2u with respect to you minus a quarter multiplied by the integral of squareu with respect to you. Let's try to integrate both of these terms. Now starting with our term here, uh, cut square u coex squared u with respect to you, then for this term, so for a quarter of the integral of square u coex square u with respect to you, again we could integrate this by substitution. So let's let some other variable t be equal to cut u. Then this would imply that the derivative of t with respect to u is negative cosex2u, so negdt is going to be equal to cosex 2 uDU. And thus that means by substitution. We can say that a quarter of the integral of cut square u cosse 2 u with respect to u is going to be equal to a negative or quarter of the integral oft 2 dt. And when we evaluate this, this is going to be a quarter or a negative quarter of t cubed divided by 3. Let me write that properly here, multiplied by T cubed divided by 3, OK. Plus a constant C which is going to give us 1/12 of cubedu plus some constant C1, OK. Now, when we have that here for our first, the first term, what about our second term, a quarter of the integral of cuts grade you with respect to you. Well, we can go ahead and evaluate that directly or integrate that directly, OK, because if we rewrite this. As Using its identity cosex 2 u minus 1 with respect to you, then when we integrate that, we get it to be a quarter of negative u, the integral of cosex squad u minus u, the integral of 1, OK, plus c2. And now that we've evaluated both, we can combine the integrals. So when we do that. OK, then. We end up with negative, let me write that here, combining both integrals. OK Then now we end up with -112 of cubed plus C1. Minus A quarter, sorry, minus 25. OK, minus a quarter of negative you minus you, OK, minus C2. And now when we do that, OK, then we'll have -1/12, OK, of cubed you plus C1 plus 24 of you plus 25 of you, OK. Minus C2. Now if we think about it here, we can combine our constants C1 and minus C2 to become some constant C. So we could rewrite this then, OK, as -1/12 of cubedu plus 24 of u, OK, plus 25 of you. Plus C, OK, where C is equal to C1 minus C2. I know that we have this in this form where we have it simplified in terms of you, we can substitute back U for 4 x. So really then this is going to be -1/12. Of cut cube of 4 X, OK. Plus a quarter of cut 4 eggs, OK. Plus a quarter of 4 x which would be x plus c. Thus, this is the integral of the cot 4 X arteries to the 4th with respect to X. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 33–52.∫ cot⁶(2x) dx

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Evaluate the integrals in Exercises 33–52.
∫ cot⁶(2x) dx