WEBVTT
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So what we want to do is prove that the
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identity arc sine of X minus one over explicit one
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secret to two times arc tangent of the square root
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of X minus pi half So something that will help
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us with this. I went ahead and define two
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functions f of X as park sign of X minus
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one over X plus one and g of X as
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two times our attention of the square root of X
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. And the reason why I want to do this
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is because both of their derivatives ends up being the
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same Attn least when we'd look, um, on
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the interval. So I should say, on zero
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to infinity, these have the same derivatives, and
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that's gonna kind of coming toe play to help us
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later. All right, Uh, so what we're
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going to want to do now is let's define 1/3
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function, and I'm gonna call this one a trucks
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, and what we're going to do is look this
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be defined as f of X minus G of X
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. Now, the reason why we want to do
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this is if we were to go ahead and take
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the derivative of H of X. This here?
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Well, give us the derivative of that. So
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we get the derivative of X minus two derivative defects
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. And the more important thing is, since both
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of these have the same derivative bowl, why subtract
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thes? I just end up with zero. Now
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, the thing that is more so important about this
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is, since this is true on it interval for
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every value of X on zero to infinity, that
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h of X is constant. So with this implies
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is that h of X is constant on zero to
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infinity. I remember this is the case. Ah
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, by one of the serums in our, um
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in this chapter now, there might be a nice
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value on zero infinity that we can plug into H
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to Europe with this constant should be end. It
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just so happens to be that each of one will
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be a nice value for us to pluck it.
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So let's go ahead and so is gonna be a
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chav. One is equal to half of one minus
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G F one and then we go had implode one
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into F, which is going to be arc sign
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of one minus one over one plus one minus two
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times are Tanja of the square root of one.
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So we end up with arc sine of well,
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one minus 10 So we just have Mark sign of
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zero minus two arc tangent of one. So arc
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sine of zero is going to be zero and then
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arc Tangent of one is pie forth. So now
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we can go ahead and simplify this down to negative
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pie half. So what we have is that h
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of X is equal to negative pie half. So
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if we go ahead and rewrite this as f of
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X minus g of X, so we have f
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of X minus G of X is able to negative
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pie half will be add g of X over.
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Now we get f of X is equal to G
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of X minus pi, huh? And then,
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if we plug in what f of X is so
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f of x waas arc sine of X minus one
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over X plus one and G of X was to
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arc tangent of the square root of X, And
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then we still have that minus pi half. And
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so this here is what we wanted to show.
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So since we finished our proof, we can parallel
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proof box and mail spotted face. So the idea
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when we're trying to do these normally is we look
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for two functions that have the same derivative. We
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can track them get zero, and then we apply
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that fear. Um, that says since the derivative
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is always zero, that means h of X has
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to be a constant on that interval.