Evaluate the integrals in Exercises 1–22.
∫₀^(π/6) 3cos⁵...

Question

Evaluate the integrals in Exercises 1–22.

∫₀^(π/6) 3cos⁵(3x) dx

Accepted Answer

Compute the integral from 0 to pi divided by 3 of 2 cosine cubed 2 XD X. We have 4 possible answers, being 3 squares of 3 divided by 8. 3 squares of 2 divided by 82 squares of 3 divided by 5, or 3 squares of 3 divided by 5. Now, how we can solve this is by using a triple angle identity, since we have cosine q. Our triple angle entity is cosine 3 theta. Equals 4 cosine cubed theta minus 3 cosine theta. We can rewrite this for cosine qs. We get 4 cosine cubed theta. This equals the 3 cosine theta. Plus cosine of 3 theta. Dividing by 4 Gives us cosine cubed theta. Equals 3 cosine theta plus cosine 3 theta divided by 4. Now, we notice that in our integral, we have 2 X in the parentheses. We can say theta equals 2X to get cosine cubed 2 X. Equals 3 cosine 2 x, plus cosine 6X divided by 4. Now we can rewrite our original integral. We have the integral 0 to 1 divided by 3 of 2 multiplied by 3, cosine 2 x plus cosine 6 X. All divided by 4. With d x at the end. And we can simplify this even further. To get Three halves. Cosine to x. Plus 1/2 cosine 6 x, all with d x. We can split this up into two integrals as well to make this easier to solve. We have the integral 0 to pi divided by 3. We have cosine to X, DX plus integral. Of 1 half cosine 6 XDX. Now let's evaluate each of these separately. We first have the integral from 0 to the pi divided by 33 halves cosine 2 X D X. Now, this is a U substitution where U equals 2 X, and DU is 2 DX. This makes DX equals 1/22 DU. We then have the integral. Of Three halves. Hoen you Multiplied by 1/2 du. And we can switch our bounds, since U is 2x, our bottom bound is 0, and our top bound is 2 pi divided by 3. Now we can simplify our integral. Integral from 0 to 2 pi divided by 3. Of 3/4 cosine u. That is going to be. 3/4 sign of u from 0 to 2 pi divided by 3. Now we can evaluate. 3/4 sin of 2 pi divided by 3. 3/4 of 0. Sin of 0 is just 0, so our second term goes away. We have 3/4 sin, 2 pi divided by 3. Based on the unit circle, sin of 2 pi divided by 3. Is the square root of 3 divided by 2, meaning we have 3 squares of 3 divided by 8. Now we can check our other integral. We have the integral. 0 to pi divided by 3. 1/2 cosine 6 XDX. Very similar to the other integral, U is 6X, DU is 6DX, which makes DX equals to 1/6 DU. We then have the integral 0. 2 pi divided by 3. But with our new bounce. Which in this case, 6 multiplied by pi divided by 3. It's just too tight. So we have 0 the 2 pi of 1/2. Cosey Multiplied by 1/6 du. We get 1/12 integral from 0 to 2 pi. Of cosine uDu, which is just. 1 twil. Sign you from 0 to 2 pi. Which in this case, sine of 2 pi and sine of 0 are just 0, meaning we have 0. The answer then is 3 square of 3 divided by 8 plus 0, or just 3 square of 3 divided by 8. This means the answer to our problem, if we look at our answers, is answer A. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Evaluate the integrals in Exercises 1–22.∫₀^(π/6) 3cos⁵(3x) dx

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Evaluate the integrals in Exercises 1–22.
∫₀^(π/6) 3cos⁵(3x) dx