2–74. Integration techniques Use the methods introduced in Sec...

Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

9. ∫ (from 0 to π/4) cos⁵ 2x sin² 2x dx

Accepted Answer

Welcome back everyone. In this problem, we want to determine the definite integral between the limits of 0 and 16th of pi assigned to the 4th of 3 X multiplied by costs to the 3 of 3 X with respect to X. Now we can determine this definite integral by using substitution. So first, let's set U to be equal to the sign of 3 X. In that case, the derivative of u with respect to X is going to be equal to 3 multiplied by the cosine of 3 X, which means then that a third of DU is going to be equal to the cosine of 3 X with respect to X. OK? So in other words, we found a way to express DU in terms of D X. Next, Taking another part of our expression, the sign to the 4th of 3 x multiplied by cost to the 3rd of 3 x, how can we break it down to incorporate the sign of 3 X and the cosine of 3 x for our expressions that we've already found here? Well, we know that the sign to the 4th of 3 x could easily be rewritten as you to the 4th. But now if we take our second term, that is cost to the 3rd of 3 X, OK, then we could rewrite it as cost squared of 3 x multiplied by the cost of 3 X. Now why is this helpful? Well, here we know, OK, that the cosine squared of 3 x by the trigonometric identity is equal to 1 minus sine squared of 3 x which we could rewrite as 1 minus U2. So now we've been able to rewrite. You, or sorry, sign 3X, OK, and DX and cost to the third of 3 X in terms of you. No, we also need to rewrite our limits. So if we think about it, if you equals 3 X, then when X equals 0, you will be equal to the sin of 0, which is 0. And when X equals a 6th of pi. Then you will be equal to 3 multiplied by 16 of pi or the sign of a 1/2 of pi, which equals 1. So we also have our limits. So now that we have all of this, we can substitute. Because now by substituting. OK. Then our integra. Is going to become a third of the integral between 0 and 1 of u to the 4th that would have been signed to the 4th of 3 x multiplied by 1 minus u squad that's core 3x multiplied by DU because remember cos 3 XD X is a third of the u and we've taken out the 3rd as our constant in our integral. No, we could take this a step further. And rewrite this, let me come down further. We could rewrite this as a 3rd, OK. Of the definite integral of you to the 4th minus you to the 6th with respect to you, OK, by expanding our bracket. No, in this case, we can integrate each term. So I leave my 1/3 of it here. I know the integral of the 4th is equal to 25th of U 5th, while the integral of u to the 6th is equal to 1/7 of u to the 7th. So we have all of this in the limits or or between the bones from 0 and 1. No, in this case we know that if we substitute 0 our entire expression will become zero, so our lower limit is 0, so we can go ahead and substitute 1. So no, that's gonna be 1/3 of 15 to 15th divided by 5 minus 17th divided by 7th. 1 to 5th and 1 to the 7th is equal to 1. So this is really 1/3 of 1/5 minus 1/7. I know if we rewrite this with the same denominator, then that's really 1/3 of 7 minus 5 divided by 35. 7 minus 5 is 2, and a third of 2/35 is equal to 2 divided by 105. Therefore, the definite integral is 2 divided by 105. Thanks a lot for watching everyone. I hope this video helped.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
9. ∫ (from 0 to π/4) cos⁵ 2x sin² 2x dx

Key Concepts

Integration Techniques

Trigonometric Identities

Definite Integrals

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