9–61. Trigonometric integrals Evaluate the following integrals...

Question

9–61. Trigonometric integrals Evaluate the following integrals.

22. ∫[π/4 to π/2] sin²(2x) cos³(2x) dx

Accepted Answer

Welcome back everyone. In this problem, we want to compute the integral of sine squared 4 X multiplied by cost cubed 4 X with respect to X between the limits of a 16th of pi and 3 of pi. Now we can compute this by using substitution. So let's start by letting you be equal to the sign of 4 X. That means that the derivative of you with respect to X is going to be equal to 4 multiplied by the cosine of 4 X, which then implies that we could also rewrite this as 1/4 of DU being equal to the cosine of 4 XD X. And that's going to come in handy because in our integral notice that we have the cosine of 4XDX included. However, this is the cosine cube.4X. So how can we eventually get it to a point to include the cosine of 4 XDX? Well, if we break it down a bit here, we could rewrite the cosine cubed of 4 XDX as the cosine squared of 4 X multiplied by the cosine of 4 XDX, OK? And in that case, we could use the trigonometric identity, the fundamental trigonometric identity, to rewrite the cosine squared of 4 X as 1 minus the sine squared of 4 X. OK? That's 1 minus the sine squared of 4 X multiplied by the cosine of 4 XDX which we could now rewrite in terms of you as 1 minus u squared multiplied by 4 of DU. OK, so we've been able to or or when we do put you into our integral, we should be able to rewrite all of the terms. Next, let's change the limits. Now changing the limits, since we have already have an expression for you in terms of X, then when X is equal to the lower limit, a 6th of pi, then you will be equal to the sign of 4 multiplied by 16th of pi or 2/3 of pi, and that is equal to 1/2 of root 3. On the other hand, when X equals 13 of pi. Then you will be equal to the sign of 4/3 of pi. Which is going to be equal to negative root 3 divided by 2. So now that we've changed the limits and we have our terms, we can go ahead. And substitute you, OK? Because now by substituting you, then our expression is now going to become the integral between a half of root 3 and a negative half of root 3 of U squared multiplied by 1 minus u squared multiplied by 1/4 of the U. And if we simplify here, if we expand our terms. Then this is going to be the integral between those limits. Of U squad minus the 4th. All divided by 4 with respect to you. Now where can we go from here? Well, let's, we can do a few things. First, we can flip the limits and change the sign. So this is really the negative integral, negative integral of route 3 divided by, between the limits of negative route 3 divided by 2 and route 3 divided by 2 of u square minus u to the 4th divided by 4 with respect to you. Next, since the function is even, then we can double the integral from 0 to a half of uh route 3, OK? So now when we do that. It's going in the next line, that's going to be a negative2 multiplied by the integral from 0 to half of root 3 of U2 minus U 4th divided by 4DU. And now, we can also take out our constant from our expression. So that's gonna be -2 divided by 4 or negative by half for integral between the limits, OK? For U squared minus U to the 4th, the U. Now this is pretty good because at this point, notice that we can evaluate our integra, OK? And now when we do that, OK, let's put back our constant negative 2. Integrating both these terms separately is going to be U cubed divided by 3, OK, minus the 5th divided by 5, OK? These are gonna be between our limits. Now, in that case, OK, then we could take this a step further. Oops, this is, uh, we could take this a step further here, should be -5, my apologies. OK, so negative 5. Now we can take it a step further and rewrite, so we have our negative half here and substitute our limits. So for the upper limit, it's going to be root 3 divided by 2 cubed divided by 3 minus root 3 divided by 2 to the 5th divided by 5. And for our lower limit, it's going to be 0 because if we substitute 0 into our expression, both of these terms would be 0. So now, let's see if we can start expanding on our terms to help us simplify, OK? So no, this is gonna be negative by half, OK, and for our first term, then we could rewrite it as 3 multiplied by route 3 divided by 8. OK. That's all divided by 3. And for a second term, it's gonna be 9 multiplied by Route 3 divided by 32, all divided by 5. And now, if we simplify that, that's gonna be negative half, multiplied by Route 3 divided by 8, because our 3 is canceled. And for the second term, that's minus 9 divided by 9 multiplied by Route 3 divided by 160. Now here, notice that we can express both of these with the common denominator of 160. So that's gonna be negative or half, multiplied by 20 root 3, divided by. Well, sorry, 20 route 3 minus 9 Route 3 all divided by 160. And now 20 route 3 minus 9 route 3 is 11 route 3 and if we distribute our negative 5 or if we multiply the result by negative 5, that's going to be -11 route 3 divided by 320. Therefore, this is going to be the integral of the sine squared 4 X multiplied by a costed 4 X with respect to X between 16th of pi and 1/3 of pi. Thanks a lot for watching everyone. I hope this video helped.

9–61. Trigonometric integrals Evaluate the following integrals.
22. ∫[π/4 to π/2] sin²(2x) cos³(2x) dx

Key Concepts

Trigonometric Identities

Integration Techniques

Definite Integrals

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