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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

32. ∫ cot⁵(3x) dx

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We are giving the integral of code to the 4th power of 2 XDX. Now, the first thing that we can do in order to simplify this integral is to use a U substitution on the innermost function. We will allow you to equal to 2 X, and that will leave us with DU is equal to 2DX. Now, we need to isolate DX in this statement, so dividing both sides of this equation by 2 will leave us with one half DU is equal to DX. Now, if we apply these substitutions, we can rewrite the integral to be the integral of cotangent to the 4th power of U, multiplied by 1/2 DU. And since one half is just a coefficient inside of the integrand, we can move that to the front of the integral as a constant. So we will have 1/2 multiplied by cotension to the fourth of U D U. Now, in order to solve this integral, we can actually use the power reduction formula for the integral of codes attention to the nth power. That identity is given as the following. If we have an integral of the form cotangent to the nth of u, then this will simplify to be negative cotangent of u minus 1 of you divided cotangent of n minus 1 of you divided by n minus 1 minus the integral of cotangent of N minus 1. Apologies minus 2 of UDU and this can be applied when n is greater than or equal to 2. Now, in our integral, cotangent is raised to the fourth power, so our value of n is going to equal to 4. And so if we were to apply the power reduction to this integral, we will be left with 1/2 multiplied by the quantity. Negative cotangent of n minus 1, which will give us 3 of u divided by n minus 1 which is 3 minus the integral of cotangent to the n minus 2 power which will square it of u d u. Now what we are going to do is we are going to distribute this one half to every quantity within the group, and so that will leave us with negative cotangent cubed of u divided by 6 minus 12 multiplied by cotangent squared of u d u. Let's go ahead and evaluate the integral on the side. Now, the integral of cotangent squared of UDU can be solved by using the Pythagorean identity for cotangent squared. We can rewrite this integral to be the integral of cosequin squared of u minus 1. And if we were to integrate both of the terms, that would leave us with the quantity negative cotangent of U minusu. So we can rewrite the integral using this value, and that will leave us with negative cotangent cubed of you divided by 6 minus 12 multiplied by the quantity, negative cotangent of U minus U, and because we were working with the general integral, we will add a general constant, C. Next, what we are going to do is we are going to distribute that negative half into the quantity that will allow us to rewrite the solution to be negative cotangent cubed of you divided by 6 plus 1/2 cotangent of you. Plus 1 half of you plus C. Now there is one more step that we need to do, and we need to resubstitute you back into terms of X. Now earlier we stated that U is equal to 2 X. So resubstituting this value for every U in the solution is going to allow us to write the solution as negative cotangent cubed of 2 X divided by 6. Plus 1/2 multiplied by cotangent of 2 X plus 1/2 multiplied by 2 X, which will just leave us with positive X. Then we will add a general constant C, and this is going to be the simplest solution to the given integral. And with that being said, the overall solution is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

9–61. Trigonometric integrals Evaluate the following integrals.
32. ∫ cot⁵(3x) dx

Key Concepts

Trigonometric Functions

Integration Techniques

Power Reduction and Trigonometric Identities

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