WEBVTT
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Let's use the formulas from exercise fifty nine. We
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took he, by definition to be tension affects over
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, too. And we use this to prove the
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other three equations. Let's quote and use that to
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evaluate this in a roll over here. So the
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first thing is, when we go ahead and take
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ecstasy over three, that's going to change the lower
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limit of integration here. So let's plug this in
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for X. See those tension of pi over six
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and that becomes route three. So that's our new
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lower limit. Similarly, playing the upper limit for
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X and then this becomes T equals changing a pirate
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for which is one that's our new upper limit.
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Then here we have a DX atop. So we
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used the ex formula over here two over one plus
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he square up top with the DC send in the
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bottom one plus, and then we have signed minus
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co sign. So we used these over here so
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sine minus coz I that will be two t trust
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he squared minus one all over one plus he square
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just adding, er subtracting here This minus co sign
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leave our numerator residence and then on the denominator,
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we can get a common denominator of one plus t
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swear. So let's pull that out and then in
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the top will have one plus T's where Plus two
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Tea plus T Square and I'm running out of room
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here. Somebody had to simplify, so we most
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upsides happened bottom of this one. Bye. One
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plus he squared. So when we do so,
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we should obtain Tootie Square plus two tea. And
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that's for the numerator within the dominant. Now cancel
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those one plus T square terms. We could also
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cancel those twos. She's the one left over three
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one. I went up top and then T square
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plus tea, and we can even pull out a
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team there. T plus one. I'm running out
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of room here. Let me go to the next
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page. Let's look at that in a grand one
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over's He t plus one. That's just one over
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T r a. O ver. T sees me
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be over t plus one and then solving for just
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multiplying both sides by this. Then I'm leader on
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the left. Go ahead and pull out a t
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three, and then we see that because there's not
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here on the left A plus B must be zero
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. The constant term on the left is a constant
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term on the right, and the left is one
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on the right. It's a so we must have
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a equals one. And then using this equation here
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, you could be his negative one. So then
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are integral becomes from roots, Rita one one over
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tea, one of the t plus one. Now
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we can evaluate that. If the second in a
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rolls bothering you because of that plus one here,
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feel free to do it yourself in the end points
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on route three and one. So now we just
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go ahead and plug those in Ellen one minus Ellen
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too, and then minus. And then here.
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So this is from plugging in won. And now
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we plug in the Route three and now we go
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ahead and cancel that that zero. We have a
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plus here, So that's a ln off three plus
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one. And then we could go ahead and combined
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these other two logarithms by multiplying them together and pulling
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out that mine design. And that's a final answer