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JH
Numerade Educator

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Problem 39 Hard Difficulty

(a) Use trigonometric substitution to verify that
$$ \displaystyle \int_0^x \sqrt{a^2 - t^2}\ dt = \frac{1}{2} a^2 \sin^{-1} \left (\frac{x}{a} \right) + \frac{1}{2} x \sqrt{a^2 - x^2} $$
(b) Use the figure to give trigonometric interpretations of both terms on the right side of the equation in part (a).

Answer

A. $\frac{1}{2} a^{2} \sin ^{-1}(x / a)+\frac{1}{2} x \sqrt{a^{2}-x^{2}}$
B. $\frac{1}{2} a^{2} \sin ^{-1}(x / a)$

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Video Transcript

for part A. Let's use a trick sub to verify the following into rule. It's on the left. We see the square root of a squared minus T square. So we should go ahead and try tricks up of the form he equals a side Ada. Let's go ahead and take the differential dt a co sign data debater. So that's our trucks up. And because here we could see that we have a definite overall our limits or zero necks with you. Go ahead and find the new limits of integration in terms of the variable data. So originally we had X equals zero. So go ahead and take t equals zero. And let's plug it into this tricks up and saw for saying data zero equals a scientist. So a non zero term we have signed equal zero, which implies that they did equal zero because remember, when we do, it shrinks up of this form a scientist. The restriction on data is that it's betweennegative pie over too, and prior to and in this case, the only solution of signing equal zero is at data equals zero. Similarly, we could find the upper limit So originally we had X So plug in T equals X Internet tricks up X equals a science data. Let's go ahead. And software data? Yeah, except yeah. And in this cues, we have Fada equals arc sine of excellent day. So I just took the sign inverse of both sides of this equation over here. And then we have our new upper limit. So let's go out and plug all this into the r a. Pinto are integral to rewrite it. So first, let's look at this radical. We have a squared minus the square, a square minus a square sign square was factor out there, eh? And then we have one minus science where that's co sign squared, and then we have this will become the absolute value of a It could be negative here. So you don't want to just write a because the square route has to be positive. So puts it absolute value. And then here we actually don't need absolute value because recall that fatal eyes betweennegative Piper to empire or two and co sign is always not negative in this region. So absolute value for co signing is not needed. Now let's go on to the next page and write out the integral. So the original integral A squared minus T squared DTS So plaguing our stuff in our new lower limit was zero still zero and the upper limit our sign Eggs over. Eh? So we'Ll have to write this every time Unfortunately And then we went head It's simplified, that radical and we had absolute value, eh? Cosign terra And then we also have dif ada Our eyes are DT which was a co sign data debater. So this right here it's coming from the DT term. Now let's go out and simplify this So the absolute value eight times, eh? We'll just be a square. You don't need absolute value here because a square is always positive or not Negative sign inverse X everday And then we're left over with co sign squared and we already pulled out the eighth square because it's just the constant Now at this, though, here we need a half angle formula that we've seen very helpful when integrating and even power of co sign or even power of science. So that's the formula. Let's go ahead and pull out That too a squared over two and then we have one plus course. I into theater. Come over here. Still have a squared over two. Let's evaluate that. Integral, Integral of one is data and in a rule of co sign to data is signed to data over to Here's our end point. So we're plugging these in for theta, so we have a squared over two sign inverse X over a and actually hear before I go ahead and plug in this value of theta. Let me go ahead and rewrite this. No. Using the double angle formula, this becomes to sign data co scientist over too. Cross off those twos and we just get signed times cause I'm so here. I have scientific co signed data, so that means I have sign sign inverse X ray times and then we have co sign so co sign sign in bors eggs over, eh? And then you could see that when we plug in data equals zero, we have zero plus I know zero, which is zero. So there's nothing to subtract, so this will be our answer. And now it's just a matter of simplifying this to make it match with the right hand side of pardon me. So now we have sorcery rival we had This was all equal to a squared over two. Just rewriting things out. Sign inverse X over a. This was data and then we had scientific co Cynthia. So this was signed. There's our data and then co sign of the same data. I can't. This was our plugging in our upper and lower limits. Now let's go ahead and evaluate this. Well, first of all, these inverse functions will just cancel each other out and we just have excelled Ray here and then we can go ahead and evaluate this So this trivia. So we want to evaluate co sign of this angle. So we have fate A equals Let me call. It's a different symbol here. We ever use data for the tricks of. So let me go ahead and use a different letter here. Let's use Alfa, which is this angle here. So we have signed in verse X ray and this means that sign of Alfa is X over. May weaken draw a triangle from this And the reason we're doing this is because we want co sign of this angle here, draw a triangle. Close it up. So here's our angles, Elsa. And then sign of Alfa is X divided by a and I want co sign of the same Alfa. So I want high pound news over a so age Bye, Petya. Great! They're a square plus X squared is a square. This means ages. Just the square root, a squared minus X squared and therefore co sign is just a chair over, eh? So let's go ahead and rewrite things over here. A squared over two sign inverse X over a Let's actually go ahead and distribute this A squared over two to both terms, No. And then plus a squared over two. Then we had sign of sign inverse of X over a that just became Excel Beret. And now we have co sign which is each divided by a and now we simplify. We see that we could cancel these a squares and we get the right hand side of a and this is what we wanted. Teo show that the integral was equivalent to this expression. So that results partner. Let's go to the next page for part B. So for part B, we like to show that the right hand side of part A. So let's rewrite that down. So first tohave part B, then we had the right hand side are hs of a is we had one half a flare. There was our angle data and then we have one half ex radical, a square minus x square And we like to show how these two terms could help us triggered a metric Lee describe the following figure that's in your textbook. So here's a rough sketch of the finger. Hey, so we have a circle of radius, eh? So it looks like here they're taking it to be a positive number and then looks like we have a sector of a circle. So, for example, this circle is centered at the origin. The circle which is coming from this equation here is equivalent to why square plus X squared equals a square. So this is just a circle. So really, I should be using a t here, not a not excellent comeback in a racist. This's a global into why square plus t squared equals a square. So this is the same circle there. This is the Red Circle. Radio is a center of the origin. Okay, so coming back then we also have the straight line segment coming straight down to x ray triangle. Okay, And then we also have our data's that are given and it looks like it's the same data. This same data that were drawn here is the same one up here, and that's just alternating interior angles. So I claim that this blue triangle is the area is given by this in the area of the center is given by this first term. So for the second term, this's just a triangle. So there we know that the area So the area of the triangle, it's just one half these times, I So here this is zero. This is the origin all the way up to X. So this is one side, so B equals X. But then we'LL also need the height chief. So since this blast segment over here in the middle is also a radius of the circle the red circle, for example, this one reason you could argue that this the length of the Saipan uses a You can also get that from the fact that state of peoples sign in verse of x ray. So this means sign of data is next divided by a So the high partners must be in. So I saw this for H. So here we have h squared. Plus exploiter equals a square. Petya. Great. There. Um so h is a squared minus X squared. So plugging these in to be in the age we have one half x room a square, minus x squared. And that matches up precisely with the second term over here. OK, so not for the first term. This is coming. This is just the area. You've seen this from geometry. This is the area of sector of circle with angle data and this happens to be our data here. So in general, the formula is one half our square data sea of a circle. You have a second center sub tended by some angle. Data in the area of the sector is given by this over here. So we are in the scenario and our data is equal to sign in verse effects over a So this is the same angle. This is this state over here. But by alternating interior er it's also this data inside the sector. So therefore we have one half is our radius radius squared times the angle theta which happens to be signed in verse of X over, eh? This means that the first term is giving you that area of the sector of the circle and that completes Barbie.