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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

16. ∫ sin²θ cos⁵θ dθ

Accepted Answer

Welcome back, everyone. Evaluate the integral. Integral of sin to the power of 4 of T, cosine to the power of 7 of TDT. So for this problem we're going to rewrite our integral as the integral of sin to the power of 4 of T multiplied by cosine to the power of 6 of T, multiplied by cosine of TDT. In other words, we are splitting off cosine, and specifically we take out cosine of T so that cosine to the power of 7 becomes cosine to the power of 6 multiplied by cosine to the power of 1. Using the properties of exponents and what we can do from here is simply introduce u substitution. Let's suppose that u equals sine of T, then the u equals the derivative of sine, which is cosine of t multiplied by the T, right? Now, what we can do is simply notice that we have cosine of TDT in our integral, and that would be DU. We have sin to the power of 4, that would be U to the power of 4. And all that we have to do is simply express cosine to the power of 6 in terms of sine. Now, first of all, cosine to the power of 6 of T can be expressed as cosine squared of T cubed. And our cosine squared is equal to 1 minus squad oft using the Pythagorean identity, right? So we get 1 minus sine squared of T. This difference is cubed, and this is equal to 1 minus u squared. Cubed Now, let's apply the formula for the cube of a difference. First of all, we have to cube our first term, so we got one cubed. Then we subtract 3 multiplied by the first term squared, so 3 multiplied by 1 squared multiplied by the second term, which is a U squared, then we. Add 3 multiplied by the first term, multiplied by the second term squared. So we use squared squared, and finally we subtract the cube of the second term, so we subtract U squared cubed. Let's go ahead and simplify. This is 1 minus. 3 U squared. Plus 3 U to the power of 4 minus u to the power of 6, and then we can substitute it for cosine to the power of 6 of T. So the integral becomes integral of you to the power of 4. Multiplied by 1 minus 3 U 2 + 3 uses the power of 4 minus the power of 6. DU And now let's distribute. We're going to get you to the power of 4 minus 3 U to the power of 6 plus. 3 to the power of 8 minus u to the power of 10 DU. And now we can integrate, so the integral of use the power of 4 is use the power of 5 divided by 5 using the power rule. Then we get minus 3. He uses the power of 7 divided by 7 once again applying the power rule. Plus 3 use to the power of nine divided by 9 once again applying the power rule and finally we can use the power rule to evaluate the integral of negative EU to the power of 10Du and that would be negative u to the power of 11 divided by 11. And let's add a constant of integration C. Now let's replace you with sign C, which gives us. Sign So the power of 5 of T divided by 5, that's our first term. Minus 3 to the power of 7 of T divided by 7. Plus 3 divided by 9 is 1/3, so we get sin to the power of 9 of T divided by 3. Mimus Sign to the power of 11 of T divided by 11. Plus C. And that'll be our final answer for this problem. Thank you for watching.

9–61. Trigonometric integrals Evaluate the following integrals.
16. ∫ sin²θ cos⁵θ dθ

Key Concepts

Trigonometric Identities

Integration Techniques

Power Reduction Formulas

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