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Evaluate the integral. $\int_{0}^{a} \frac{d x}{\…

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Problem 6 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_0^3 \frac{x}{\sqrt{36 - x^2}}\ dx $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

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Video Transcript

here we have a definite nibble from Sierra three X over the square root of thirty six, minus x squared. Since our denominator is of the form ace flirt minus X squared. We should use the tricks of of the form X equals a sign data. Here we see that a equal six. So we should take X to be six in data then D X is six co signed data data. So notice here we have. Ah, definitely a rule. So we have two options on how to proceed. The first option is to find the new limits of integration in terms of data. And if you go that road, you won't have to draw the triangle and you won't have to back substitute because there won't be any need to go back into the variable X. So I prefer that option. The second option is to draw, evaluate, integral, draw the triangle back, substitute back into the variable X, and then you could plug in zero and three is your ex values into the anti derivatives, so that will require the triangle if you want to go that way. So in this problem, let's see if we could find the new limits of integration. So here noticed that from the textbook. When we make this trick substitution, we always take fader to be between negative power too. And powerful too. This will help you find data. So the lower limit before was zero. So if we plug in X equals zero into this equation up here, excuse me into this equation down here, we could solve her data. So we have zero equals six i data. So this means signed data equals zero and sense data has to fly in this interval. There's only one solution, and that's exactly data equal zero. So that's it. One of the new lower limit of integration. The upper limit was ex people story. So plugging X equal three into this equation again, we have three equals. Six idea, so scientific is a half. And then, in this interval from negative Piper Tuna Piper to the only time data is a sign Data's a half is when data is five or six. So that's the new upper limit. So here one more observation. We have this radical in the denominators. Let's quote and simplify that First thirty six minus X squared is thirty six minus thirty six Sign square. We can pull out a thirty six out of the radical that become six. Then we have one minus sign square. So this is equal to six square rule of co signs where which is equal to six co signed data. Come over here and write that six co signed data so that that's our new denominator. So now we have in a rule zero to power six x and the numerator. So it's six nine data DX, which was six coastline data. And then we have six co signed data on the bottom. That's coming from the denominator that we simplified. We could go out and cancel off these six co signs. Pull out the six. So we have six Integral Sierra five or six signed data. So this is negative. Six. Let me write it this way. Six anti derivative of sinus negative co sign and then we have our end points zero, five or six. So plugging in those end points. We have negative six times radical three over too plus six, which we can write as negative. Three. Room three plus six. And there's our answer

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Problem

Evaluate the integral. $\int_{0}^{a} \frac{d x}{\…

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Related Courses

Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 7

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Anna Marie Vagnozzi

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Anna Marie Vagnozzi

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Anna Marie Vagnozzi

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals