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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

47. ∫ (csc⁴x)/(cot²x) dx

Accepted Answer

Hello. In this video, we are going to be computing the following integral. We are given the integral of cosequent to the 6th power divided by coage into the 4th power of X. Now, the first thing that we want to do is we want to try to reduce the exponents as much as possible so that we can go ahead and apply trigonometric into identities. Now, let's go ahead and try to represent everything in terms of cotangent. Now, the first thing we can do is we can reduce cosequin to the 6th power as a product of Cosequin squared and cosequant to the 4th power. By separating these values, we can rewrite the integral as the integral of cosequent squad of X multiplied by the quantity cosequent to the 4th power of X, divided by cotangent to the 4th power of XD X. Now, the next thing we're going to do is we are going to rewrite Cosequant into the form of a squared power. We can represent Kosekin to the 4th power as Kossein squared raise to the power of 2. So we can rewrite the integral as cosequent squared of X multiplied by the quantity cosequent squared of X raised to the power of 2, and this will be divided by cotangent to the 4th power of X. Finally, we can go ahead and apply a trigonometric identity to cosin squared to represent it in terms of cotangent. We can rewrite the integral to be cosequin squared of X multiplied by the quantity 1 plus cotangent squared X, raise the power of 2, divided by cotangent to the 4th power of X. Now what we want to do is we want to go ahead and simplify the integral by using a U substitution. Let's go ahead and allow you to equal to the more problematic term, which is this cotangent in the denominator. So, if we allow U to equal to cotangent of X, then D U is equal to negative cosequent squared XDX. If we take a look at our integrand, we have cosequent squared of X, DX and the integrand, but we don't have a negative sign. So if we were to just divide both sides of this equation by -1, We will be left with negative DU is equal to positive cosequin squared XD X. So the cosequin squad of XDX in our integrad is going to substitute with negative DU and every cotangent we have will simplify with you. So applying these substitutions allows us to simplify the integral to be the integral of the quantity 1 plus U squared, raise the power of 2, divided by U DU, and now what we're going to do is we're going to simplify this integral, oh sorry, I forgot the negative sign because of negative DU. We'll put that to the front of the integral, and now what we're going to do is we're going to simplify by using algebraic methods. So if we were to expand the numerator, we will be left with the negative integral of 1 + 2 u 2+ u to the 4th power. Raised divided by you to the 4th power do you. Then we are going to split this into a sum of fractions, leaving us with the negative integral of 1 divided by u to the fourth power plus 2 u squad divided by u to the 4th power plus the 4th power divided by u to the 4th power DU. Further simplifying will leave us with the negative integral of u to the 4th power plus 2u to the negative 2 power plus 1 u. All we're going to do now is we're going to take the antiderivative of each term within this integral. By taking the antiderivative, the integral will evaluate to be the negative value of negative 1 divided by 3 u cubed minus 2 divided by U. Plus you, and because we were working with the general integral, we will add a general constant c. Now, the last thing to do here is we are going to first distribute that -1, which is going to give us 1/3 U cubed minus, plus 2 divided by U minus U plus C. and now we're going to resubstitute you back in terms of X. Earlier we stated that U is equal to cotangent of X. So by plugging in cotangent of X for every U term, we are left with 1 divided by 3 multiplied by cotangent cubed of X plus 2 divided by cotangent of X. Minus cotangent of X plus 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 bed. 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.
47. ∫ (csc⁴x)/(cot²x) dx

Key Concepts

Trigonometric Functions

Integration Techniques

Trigonometric Identities

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