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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

50. ∫ csc¹⁰x cot³x dx

Accepted Answer

Hello. In this video, we are going to be evaluating the given integral. We are given the integral of cose into the 6th power of X, multiplied by cotangent cubed of X D X. Now, in order to approach this problem, the first thing we're going to try is we're going to try a U substitution. If we allow U to equal to cosequent of X, then DU is going to equal to negative cosequent of X multiplied by cotangent of XD X. And if we were to move the negative sign to the left hand side, we have negative DU is equal to cosequent of X, cotangent of XDX. Now, we don't have cosequant multiplied by cotangent DX in the integrand, but we can rewrite the integrand to have these terms. First, let's go ahead and pull a multiple of cosequant and cotangent from the integrand. That will allow us to rewrite the integral to be the the integral of cosequent to the 5th power of X multiplied by cotangent squared of X, and this will be multiplied by cosequent of X cotangent of X. D X So, we have the term we need to apply the U substitution, but let's go ahead and represent cotangent squared of X into terms of co-sequent. So, we can go ahead and apply the Pythagorean identity for cotangentive X. The Pythagorean identity will allow us to rewrite cotangent squared X to be cosequin squared X minus 1. So we can further rewrite the integral to be the integral of cosequant to the 5th power of X, multiplied by the quantity cosequent squared of X. -1 Multiplied by cosequent of X, multiplied by cotangent of XD X. And now we can apply our U substitution. Here, cosequant multiplied by cotangent DX will substitute with negative DU, and every cosequant term here is going to substitute with u. So that will simplify the integral to be the integral of u to the 5th power multiplied by U2 minus 1, multiplied by negative dU. And if we were to move that negative to the front of the integral, then we have the negative integral of the product of the U quantities to you. Now what we are going to do is we are going to simplify algebraically. We are going to distribute you to the 5th power into you quad minus 1, leaving us with the negative integral. Of you to the 7th power minus you to the 5th d you. By taking the antiderivative of each term in the integrand, we are left with -1 multiplied by 1/8, power, minus 16 U 6 power, and because we were working with a general integral, we will add a general constant C. Distributing this -1 will leave us with negative 1/8 U8 power, plus 1/6 U 6 power plus C. And what we are going to do now is resubstitute you back in the terms of X. Earlier we stated that U is equal to cosequant of X, and so that is going to leave us with negative 1/8 multiplied by cosequent to the 8th power of X plus 1/6 multiplied by cosequent to the 6th power of X plus C. and this is going to be the general solution to the given integral. And with that being said, the overall solution is going to be D. 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.
50. ∫ csc¹⁰x cot³x dx

Key Concepts

Trigonometric Functions

Integration Techniques

Pythagorean Identities

Watch next