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.

51. ∫ (from 0 to π/4) sin⁵(4θ) dθ

Accepted Answer

Welcome back, everyone. Evaluate the integral, integral from 0 to pi divided by 3 of sine race to the power of 5 of 3 XDX. So first of all, let's consider our integrant. We have sin raised to the power of 5 of 3 X. What we are going to do to integrate this function is simply apply several manipulations. We can rewrite sine raise to the power of 5 S. Sine raise to the power of 4 of 3 X multiplied by a sine of 3 x using the properties of exponents. Now, let's consider sine raises to the power of 4. We can rewrite it as sine squared of 3 X. Squared And then we're multiplying by a sign of 3 X. And now let's recall the Pythagorean identity. It essentially states that sin squared can be rewritten as 1 minus cosine squared, so we get 1 minus cosine squared of 3 X squared. Multiplied by a sine of 3 X. And now we're going to introduce a use substitution. Let's suppose that you. Equals cosine of 3 x. Then the derivative d u divided by the X is equal to. The derivative of cosine of 3 X, which is negatives of 3 X multiplied by the derivative of 3 X, which is 3. We can now express sine of 3 XDX from this relationship. This is going to be useful because we have sin of 3 X, and then we're going to have the X within the integra, right? So sin 3XDX is going to be equal to DU divided by -3 or 13 DU. Our final step is to evaluate the new limits of integration. U1 is going to be equal to cosine of. 3 multiplied by 0, so that's cosine of 0, which is equal to 1, and the upper limit of integration U2 is going to be equal to cosine of 3 multiplied by pi divided by 3, right, which is cosine of pi. And it is equal to -1. So now we can rewrite our integral as the integral from 1 to -1. Of 1 minus. You squared squared. And then we have sin of 3 X multiplied by DX, which is 1/3 DU. We're just going to write the EU and we're going to factor out -13. Well then, so we have our integral in terms of you. Now, let's apply the reverse interval property. We can essentially change the negative sign with the positive sign, so -1/3 becomes positive 1/3, and we're going to swap our limits from -1 to 1. And we are integrating 1 minus u2 squad which can be written as 1 minus 2u2 plus a u raises the power of 4DU. We are basically squaring right. And now let's integrate. We have 1/3. In pri C we can integrate one which is simply equal to you. We are then subtracting 2 multiplied by the integral of U2, which is U cubed divided by 3 using the power rule. And then we're adding the integral of U to the power of 4, which is U to the power of 5 divided by 5. Once again, using The power rule. And now let's evaluate the result from -1 to 1. First of all, we're factoring out 1/3 and imprec we are replacing every u with 1. So we get 1 minus 2/3 + 1/5. And we are subtracting the result at the lower bound 1/3 in -1 minus. 2 multiplied by -1 cubed divided by 3, so we're basically adding. 2/3, and we're adding -1 raises to the power of 5 divided by 5, which essentially means that we are now subtracting 1/5. Now let's take out 1/3 in pregnancies, we're going to have 1 minus 2/3 plus 1/5. And now because we are subtracting, we are going to take every term from the factor at the end with the negative sign, right? So we are subtracting -1, meaning we're adding one, we're subtracting. 2/3 And we're subtracting -1/5, meaning we're adding, right? So now let's simplify the results. This is equal to 1/3. In prics we have 1 + 1 which is 2, 2/3 minus 2/3 is. 4/3. And then + 15th and plus 1/5 is 2/5. We can now identify the common denominator. Which is going to be 15. So we end up with 1/3 multiplied by 2 can be written as. 30 divided by 15. -4/3 can be written as -20 divided by 15, and 2/5 can be written as. 6 divided by 15. So now we have 30 minus 20, which is 10 and 10 plus 6 is 16, right? So we got 16. Divided by 3 multiplied by 15, which is equal to 16 divided by 45. That's our final answer and thank you for watching.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
51. ∫ (from 0 to π/4) sin⁵(4θ) dθ

Key Concepts

Integration Techniques

Trigonometric Identities

Definite Integrals

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