integrate-each-of-the-given-functions-int-frac03-d-ssqrt2-s-s2-3

Question

Integrate each of the given functions.
$$\int \frac{0.3 d s}{\sqrt{2 s-s^{2}}}$$

Tyler Moulton · Accepted Answer

we want to integrate the following expression. The integral of 0.3 Ds over route to s minus X squared. Or as I&#x27;ve written on the right edge grill of route three Ds Or rather 30.3 Ds over route one minus x minus one square. This question is testing our knowledge of integration techniques and single variable calculus in particular is testing our newfound knowledge of inverse trig integral on the left. One integral. Do you over a squared minus u squared? Is arsenio april see on the right to integral to you over a squared plus b squared is one of our our changing over a policy. Thus we see are integral cause it has a square root in the denominator is of the form one, which means Saw, we must now figure out what you do you and they are the way I&#x27;ve written on the right and it&#x27;s easy to see that you is that&#x27;s minus one, do you? And DS. And a point is it is one that should be an extra factor of 10.3 and are integral must carry over to the solution, which means we have solution 0.3 integral DS over one minus x minus one squared is three Or rather .3 times the arc tangent. Already arc sine I was at minus one plus E. That is our solution is boxed here are highlighted .3 Arc sine -1, S -1 Plus C.

Foster Wisusik · Answer

Okay, This question wants us to find this and cider. So to do this, we really only have one option for you. So let&#x27;s pick the whole thing under the square root. It&#x27;s have you equal to three minus two s. So d&#x27;you is just negative to d s. But we only have one d s inner integral. So to get that let&#x27;s just divide both sides by negative to to get negative, do you? Where to equals D s. So now we have stuff that we can plug in. So are integral is just integral of square You times negative. Do you over too or negative 1/2 times the integral of you to the 1/2 do you? And this is just negative. 1/2 times our anti derivative, which is 2/3 you to the three halves plus c or multiplying this constance out. We get negative 1/3 you to the three halves plus c. So now we just gotta plug back in for you, which we said was three minus two s. And this is our anti derivatives

Matt Just · Answer

Okay, So another definite integral from negative to the negative one of this function. Negative to over tea plus three e to the zero point three team Dean. So anti derivative. So? Well, just remember that. Um okay, so, So anti derivative of one of her t is log absolute value of tea, since it&#x27;s going to be negative too. Log Absolutely. T plus. Okay, well, this is going to be three eating a zero point three t, but I might have to divide by this constant factor. I just think when you take the derivative, you multiplied by zero point three by the chain roll you were just going to divide. So I have three divided by zero point three. Me too. The zero point three t evaluated from negative to donate one. Okay. So again, we can just actually evaluate each one of these turns at the end points and take the difference. So and you have two times log to minus log one and notice that I&#x27;m applying the absolute value to negative one and and negative, too, and actually ended their records. So long. Absolute value of one minus long. Absolute values too. Plus Okay, three overs your point three. That&#x27;s one over&#x27;s your point. One or one over one tenth. Let&#x27;s just ten. And then we have into the minus zero point three. Brian&#x27;s A to the minus, sirrah point six. Okay? And so that&#x27;s good. I mean, one thing we can do. Okay, Log of one zero. So this is just to sow the negatives. Cancel natural, too. Plus ten times this difference. At this point, we can probably just plug this end and what we get Approximately three point three zero six.

Xiaomeng Zhang · Answer

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Doruk Isik · Answer

this problem you&#x27;re asked to, lady, Follow Inter Go. Um, first of all, we don&#x27;t have screwed up as there&#x27;s nothing about std on. Um, how are all right knowing this. Now, let&#x27;s do like the first Dermot first on the Newman Energy numerator by espirit and rewrite this integral Ben as one plus, let&#x27;s do to say important 2nd 1 So we buy the second term in new berated by a spread, we get a CZ two b negative three over two he s and the entire narrative or one will be s. And the entire narrative or Esther negatively or Jew will be as to the native would help you find. Or now we can run this oneness as minus two. Delighted by a spirit of s um what&#x27;s it like this? That one. Unless he really is at a spirit of two people, that&#x27;s that&#x27;s occurred to you find it to be spreading up to minus two or 2 to 1 or power minus evil. That that one, we have one minus two. From this, we find the answer to be screwed up too. Waas one minus to the trioval.

Ian Maurer · Answer

Okay, so we&#x27;ve got an integral from one to the square root of two we&#x27;ve got s squared. Plus route s over s squared. That&#x27;s square, plus route ass. Whoever s squared T s. Do you want to put this up? Feels like you&#x27;re distributing. The denominator s squared over s squared. I&#x27;m gonna go ahead and switch a square root to 1/2 power while we&#x27;re at it. D s like that, we can, uh, further simplify that. Still integral for one of the spirit, too. That&#x27;s squared. Divided by a spirit. It&#x27;s one as to the 1/2 divided by s to the two you&#x27;re subtracting 1/2 minus two. And so that&#x27;s gonna be s to the negative three halves because you&#x27;re subtracting before has essentially Now you&#x27;re ready to do your anti derivatives anti derivative of one will be s anti derivative of the second piece will be asked in the negative 1/2 because he add one and then you divide and that&#x27;s evaluated on our two bounds. Want to square it too? We plug in the top value first. So you&#x27;re square. Read it too. Plus query to to to the negative 1/2 divided by negative 1/2. Then we plug in one one plus one to the negative. 1/2 over. Negative 1/2. Let&#x27;s see what we can do to make that better. Let&#x27;s see, dividing by a negative 1/2 is the same. Is multiplying negative too. So I&#x27;m looking at square root of two minus two. And now, if I raise something to the negative 1/2 it goes on the bottom. So I have, uh, to to the one forth, because it was a square root and another square root. It doesn&#x27;t look too nice, but that&#x27;s okay. Minus the other side is one minus two. Uh, great. So I guess our answer is square root of two minus to over two to the 14 minus negative ones. I could destroy it. Plus one. I don&#x27;t really know exactly how your instructor would like it. I&#x27;ll just show you one other option. These air, these last two expressions are exactly equivalent. You could have gone like this and had it be the fourth through of two close one. Or I guess you could have done to to the 1/2 minus two times two to the fourth plus one. So I don&#x27;t know exactly which one

Problem

Integrate each of the given functions. $$\int_{\p…

Problem 22 Easy Difficulty

Integrate each of the given functions.
$$\int \frac{0.3 d s}{\sqrt{2 s-s^{2}}}$$

Answer

$0.3 \sin ^{-1}(s-1)+C$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Heather Z.

Caleb E.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$0.3 \sin ^{-1}(s-1)+C$

Basic Technical Mathematics with Calculus

Grace H.

Heather Z.

Caleb E.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Heather Z.

Caleb E.

Kristen K.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus