7–84. Evaluate the following integrals.
41. ∫ cot^(3/2)x...

Question

7–84. Evaluate the following integrals.

41. ∫ cot^(3/2)x · csc⁴x dx

Accepted Answer

Welcome back, everyone. Evaluate the integral integral of cotangent to the power of 1/2 of X, co-sequence to the power of 4 of XD X. For this problem we're going to use the U substitution. Let's suppose that U equals cotangent of X. Our next step is to identify the derivative DU divided by D X, which is the derivative of cotangent, and that's negative cos 2 X. We have to notice that within our integral. Cotangent to the power of 1/2 of X cosequent to the power of 4 of XD X we have cosequent to the power of 4, right? What we can do is simply rewrite it as cosequent squad of X multiplied by cosequant squad of X followed by D X. So now what we can do is simply rearrange our derivative calculated previously and show that coc can 2 X. The X is equal to. Negative DU, right? So we can mark that part within our integral coc squared XDX is going to be negative DU. We know that cotangent to the power of 1/2 of X is going to be equal to use the power of 1/2. So what we want to do is simply focus on cos squad of X. Let's recall trigonometry, and let's remember that cosequent squared X is equal to 1 plus cotangent squad of X. Or basically 1 plus U squared. Meaning we can rewrite our integral as the integral of use to the power of 1/2 multiplied by cosquad, which is 1 + U2. Multiplied by negative DU, so we can just write DU and factor out the negative sign. And now we're going to evaluate this integral. Let's label it as I. By distributing So we get negative. Integral of you to the power of one half do you? Plus Integral, use the power of 2 + 1/2, right? So that'd be use the power of 5 halves, the you. And now we are ready to evaluate each integral using the power rules so we get negative. Imparencies use the power of three halves divided by 3 halves or simply multiplied by 2/3, right? Plus you the power of 5 halves plus 1 is 7 halves, and then we're dividing by 7 halves or multiplying by 2/7s. And finally we want to add a constant of integration C because this is an indefinite integral. And now let's substitute back, so we got 2/3. U is cotangent, so we get cotangent to the power of three halves. Of X Minus because we have a negative sign in front 2/7 cotangent to the power of seven halves of X. Plus see, which is our final answer for this problem. Thank you for watching.

7–84. Evaluate the following integrals.
41. ∫ cot^(3/2)x · csc⁴x dx

Key Concepts

Integration Techniques

Trigonometric Identities

Improper Integrals

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