7–84. Evaluate the following integrals.
30. ∫ from 5/2 t...

Question

7–84. Evaluate the following integrals.

30. ∫ from 5/2 to 5√3/2 [1 / (v² √(25 - v²))] dv

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate the integral of 1 divided by 2 squared multiplied by the square root of 16 minus W squared with respect to W between the limits of 4/3 and 4/3 of route 2. Now, let's try to evaluate this integral using substitution. First, let's let W be equal to 4 sine theta, OK? So we have an expression for W. Now if we differentiate W with respect to theta. Then it's going to be equal to forecast data, OK, which means then that DW. Is going to be equal to 4 theta D theta. So in essence, then we found an expression for W and DW to aid in our substitution. The final thing we need to do is to get an expression for the square root of 16 minus W squad in terms of theta. Well, since we already have an expression for W, that means the square root. OK, of 16 minus W2 is going to be equal to the square root of 16 minus 4 sin theta squared, OK, which, when we expand that, that's going to be the square root of 16 minus 1616 square theta. And if we factor out 16 from our expression, that's gonna be the square root of 16 multiplied by the square root of 1 minus sine square theta. And if we think about it, the square root of 16 is 4, while 1 by the trigonometric identity, 1 minus sine squared theta is squad theta, and the square root of cost grade theta is simply cost theta. So the square root of 16 minus W squared would be equal to 4 cast theta in terms of theta. So now that we have an expression in terms of theta for that for DW and for W, then we can go ahead and substitute but wait, remember that our limits are with respect to W. so we'll also need to change our limits for it to be with respect to theta, OK? So changing the limits if we think about it here. Then starting with our lower limit. If W, and let's call it W1 for a lower limit, is equal to 4/3, OK, then that means 4th sin theta is going to be equal to 4/3 because remember W equals 4 sin theta. And now if we cancel 4 from both sides, that means the sin of theta 1 or lower limit will be equal to 1/3, OK. Which means that the arcs, that means that theta one, OK. It is gonna be equal to the arc sign of 1/3. So this will be our corresponding lower limit in terms of theta. On the other hand, OK, if W is equal to 4/3 of route 2, OK. Then, like we deduced before, that means 4 sine theta will be equal to 4/3 of route 2, which means that the sine of theta 2 will be equal to root 2 divided by 3, and thus this implies that theta 2 or upper limit will be equal to the arc sign. Of rode 2 divided by 3. So now we also have our new upper and lower limits for theta. So with all this information. Then substituting in our original integral. For theta, OK. That means no, it's going to become the integral between the limits of arc sine of 1/3 and arcs sign of route 2 divided by 3, OK, of 1 divided by W squared. OK. So that's gonna be 4 sin theta squared or 16 sin squared theta, OK. Multiplied by the square root of 16 minus W2 which we found to be forecast theta. And that will be res with respect, that will be multiplied, sorry, by forecast theta D theta to replace DW. Now, this is pretty helpful because here, you can probably see why we not chose 4 sine theta as well, because forecast the cancels in the numerator and the denominator, OK? And if we were to factor out 1/16th or a constant, Then we'll get this to be equal to 1/16 of the integral between our limits for theta. Let me just write them in here, OK. OK, of one divided by Sanskrit theater or Coex theater based on uh the reciprocal definition, that's the reciprocal of Sanskrit theater with respect to theater, OK? Now, if we go ahead here, if we integrate coex created theta, it equals negative cut theta. So then this is gonna be 1/16 of negative cut theta, OK, between our limits of arcs 1/3 and the arcs sign of route 2 divided by 3. Now, where do we go from here? Well, we can evaluate our core tangent with those limits, OK? So starting with the cotangent. Of the arc sign of route 2 divided by 3. OK. By definition, we know that is gonna be equal to the square root of 32 minus route 22, OK? All divided by route 2. 3 squared is 9, Route 2 squared is 2, and 9 minus 2 is 7. So that's route 7 divided by route 2. On the other hand, the cotangent of the arcine of 1/3, OK. It's gonna be equal to the square root of 32 minus 12 divided by 1. And then we simplify that would be equal to root 8 or to root 2. So now that we have the cotangent values for both of those limits, then we can put them back into our equation. OK, our expression, and that is gonna then be 1/16 multiplied by the upper limit, negative route 7 divided by route 2. Plus or lower, or minus, sorry, or a lower limit, so that's minus the negative value of 2 root 2. So that's gonna be plus 2 root 2, OK. And no, if we try to simplify here, if we combine both of the terms inside our bracket, that's gonna be 1/16 of 4 minus root 7 divided by route 2, and if we multiply through by 1/16, this will be equal to 4 minus route 7 divided by 16 route 2. Thus this is going to be the integral of. 1 divided by W squared multiplied by the square of 16 minus W2 with respect to W between the limits of 4/3 and 4/3 of root 2. Thanks a lot for watching everyone. I hope this video helped.

7–84. Evaluate the following integrals.
30. ∫ from 5/2 to 5√3/2 [1 / (v² √(25 - v²))] dv

Key Concepts

Definite Integrals

Integration Techniques

Fundamental Theorem of Calculus

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