Evaluate the integrals in Exercises 1–22.
∫₀^(π/2) sin²(...

Question

Evaluate the integrals in Exercises 1–22.

∫₀^(π/2) sin²(2θ) cos³(2θ) dθ

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the definite integral. The integral evaluated from 0 to pi divided by 2 of sin to the power 3 of 2 x multiplied by cosine squared of 2 x dx. OK, so it appears for this particular prom, we're ultimately asked to evaluate the definite integral that is provided to us by the prom itself. So now that we know that we're ultimately trying to evaluate this particular integral. Our first step that we need to take based on our prior knowledge is we need to rewrite the integrand. We also need to note that. Sign to the power of 2 of 2 X, we need to note that this will be equal to 1 minus cosine squared. Of 2 X. Fantastic. So with that said, we now can safely write that the interval evaluated from 0 to pi divided by 2 of sin to the power of 3 of 2 x multiplied by cosine to the power of 2 of 2 x dx. Will be equal to drum roll, the integral evaluated from 0 toi divided by 2 of sin to the power 2 of 2 x multiplied by sin of 2x multiplied by cosine of, or specifically cosine to the power of 2 of 2x. D X Will be All equal to drumroll, the integral evaluated from 0 to pi divided by 21 minus cosine square of 2x multiplied by sin of 2x multiplied by cosine squared. Of 2 X. DX is going to be all equal to the integral evaluated from 0 to pi divided by 2 parentheses of cosine square of 2 x minus cosine to the power of 42 x multiplied by sin of 2 x dx. Fantastic. So now what we need to do is we need to recall and use the substitution method, and let us state that you for this particular prompt will be equal to cosine of 2 x. Then that will mean that DU by DX will be equal to -2 multiplied by sin of. 2 x, which will mean we could write that 1/2 multiplied by duu will be equal to sin of 2 x. DX fantastic. So now we need to rewrite our limits. So as X is equal to 0, that will mean that You will be equal to. Parentheses of, so once again, you will be equal to, you don't have to write in parentheses, but you could say U is equal to cosine of 2 multiplied by 0. These we're plugging in a value of 0 in for x. So that will mean that cosine of 2 multiplied by 0 will be equal to cosine of 0, which is simply equal to 1. And X is equal to pi divided by 2, which will mean that you will be equal to cosine of 2 multiplied by parentheses of pi divided by 2, which will be equal to cosine of pi, which will be equal to -1. And we know that these values based on our knowledge of trigonometry. So, moving right along here, we now need to substitute back into our integral. So we need to take the integral evaluated from 0 to pi divided by 2 of parentheses of cosine square of 2x minus cosine to the power of 42 X. Multiplied by, so we need to multiply this by. Sign up to XDX. And this will be. So drumroll, this will be equal to the integral once we, since we've rewritten our limits, this will be equal to the integral evaluated from 1 to -1 of u2 minus u 4 parentheses multiplied by 1/2 multiplied by du in parentheses. Fantastic. We now need to recall that for any integrable F function F on the closed interval square brackets of A, B, that will mean that, let's make a note that the integral evaluated from A to B of FU multiplied by DU. Will be equal to negative the integral evaluated from B to A of FUDU. Fantastic. And once again, we need to note that this means that we need to note, once again, based on our prior knowledge that for any integral function F on the closed interval A, B, that will mean we can write, we can apply this methodology, this piece of information for our integral. So, With that said, Moving right along here, we're on a roll. We can therefore, based on this knowledge, we could go ahead and write that the interval evaluated from 1 to -1 of U 2 minus u 4 in parentheses multiplied by 1 divided by 2 multiplied by DU will be equal to 1 divided by 2 multiplied by the integral. The integral from -1 to1 of u2 minus u 4 in parentheses. Fantastic. So now, we need to integrate by recalling and using the power rule, which states that the integral of u to the power of NDU is equal to u n + 1 divided by n + 1 + C. Where we need to note that N, this integer value n cannot equal -1. So when we recall and apply the power rule, we will be able to write that. 1 divided by 2 multiplied by the integral evaluated from -1 to 1 of u to the 2 minus u 4 in parentheses DU. Will be equal to 1/2 multiplied by u to the power of 3 divided by 3 minus u to the 5 divided by 5 evaluated from -1 to 1. And this will be all equal to it's gonna be equal to 1 divided by 2 multiplied by square brackets of parentheses 1 to the power of 3 divided by 3. -1 5 divided by 5 in parentheses minus of 13 divided by 3 minus 15 divided by 5. In parentheses, and don't forget your square bracket, and when we evaluate this expression, when we plug in this big old long expression into our calculator, and we write it in our most simplified fractional form, we will find that it's simply equal to 2 divided by 15, and that is our final answer. Hooray, we did it. So, going back to the top to to go over everything that we've learned to To know what our final answer will be, we can, or recap would be the word. To recap what we've learned, we can say without a doubt that our final answer for the evaluated definite integral without a doubt will be 2 divided by 15, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the integrals in Exercises 1–22.∫₀^(π/2) sin²(2θ) cos³(2θ) dθ

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Evaluate the integrals in Exercises 1–22.
∫₀^(π/2) sin²(2θ) cos³(2θ) dθ