Use the substitutions in Equations (1)–(4) to evaluate the int...

Question

Use the substitutions in Equations (1)–(4) to evaluate the integrals in Exercises 33–40. Integrals like these arise in calculating the average angular velocity of the output shaft of a universal joint when the input and output shafts are not aligned.
∫(from π/2 to 2π/3) cos θ dθ / (sin θ cos θ + sin θ)

∫(from π/2 to 2π/3) cos θ dθ / (sin θ cos θ + sin θ)

Accepted Answer

Welcome back everyone. In the analysis of a spatial linkage, the instantaneous angular rate Rx of one arm relative to another is given by r x equals sin x divided by 1 plus cos x. The net angular change as x goes from 1/3 of pi to 12 of pi can be found by evaluating the integral between a half of pi and 1/3 of pi of sin x divided by 1 plus cos x with respect to x. Evaluate this integral by using the Westrass substitution where z equals the tangent of a half of x plus x equals 1 minus z 2 divided by 1 + z2. Sin x equals 2 z divided by 1 + z 2 and dx equals 2dz divided by 1 + z 2. A says the answer is the natural log of 3 halves. B says it's the natural log of 5 halves. C says it's root 3 divided by 2, and D says it's a half. Now how are we going to evaluate this integral by using the Weierstrass substitution? Well, if we let our integral bei, OK, so i equals the integral between 1/3 of pi and 1/2 of pi of sin x divided by 1 plus cost x with respect to x. Then we use the Westras substitution by evaluating this integral with respect to z. So in other words, we're substituting z into our integral. Now to do that, so far we know what z is. We know what the cosine of x and the sin of x and the dx is or are in terms of z. So all we need to do now is to transform our limits based on the variable z, and then we can solve. So let's focus on our limits. Starting with our lower limit. When x equals 1/3 of pi, then z will be equal to the tangent of pi divided by 6, OK, because that would be a half of a third of pi, which equals 1 divided by root 3. When x equals 1/2 of pi. Then z would be equal to the tangent of 1/4 of pi, which is equal to 1. So now we can substitute sin x, class x, and dx into our integral because this no means that i is going to be equal to the integral between z1 and z2. Of 2 Z divided by 1 + Z2, OK, that's sin x. Divided by 1 + 1 minus Z2 divided by 1 + Z2, that's class x multiplied by 2 DZ divided by 1 + Z2, that's dx. Now, to take it a step further, let's try to simplify. Let's start by simplifying our denominator. Now notice here, let me put this in a box beside our work, OK. Notice that 1 +, OK, 1 minus Z2 divided by 1 + Z2 is equal to 1 + Z2. Plus 1 minus z2 divided by 1 + z2 when we express it in terms of one denominator and this simplifies the 2 divided by 1 + z2 so now we can put that into our integral we can rewrite it then as the integral between z1 and the Z2. OK. Of 2 Z divided by 1 + Z2 divided by 2 divided by 1 + Z2, OK, multiplied by 2DZ divided by 1 + Z2. And now it simplifies, OK, because this all of this term will simplify it to z. So really this is z multiplied by 2 dz divided by 1 plus z2, OK? And again let's just put our limits z1 and the z2 here, and we can take this a step further to rewrite it. OK, as the integral of 2 z divided by 1 + z 2 with respect to z. No, we're almost there. All we need to do now is to figure out this integral. Let's do it by using some substitution. So let's let you be equal to 1 + Z2. Well, in that case, that means the derivative of u with respect to z would be equal to 2 z or d u then would be equal to 2 z dz. So really our integral I, OK, is going to become our integral. OK, of one divided by you, OK. DU Now do we know anything about this integral? Well, yes, by definition, we know that this is equal to the natural log or ln of the absolute value of u plus c. Which means then this will be LN of 1 + Z2 plus C. OK, now all you need to do to solve is to evaluate from Z1, which equals 1 divided by root 3, to Z2, which equals 1. So no, this means then that I is going to be equal to. Ellen of. 1 + Z2, OK. Between the bones are the limits of 1 divided by route 3 and 1. So that's gonna be LN of 1 + 12 minus LN 1 + 1 divided by root 32. And now when we simplify here, this becomes LN2 minus LN of 4/3. Based on what we know about logs, we could or the properties of logs, we could rewrite this as the LN of 2 divided by 4/3. Which is going to be equal to the ele of 3 halves. Thus, this is going to be the value of our integral. If we look back at our answer choices, that means that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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