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When we're integrating, we need
to be able to recognize standard
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forms and use them immediately.
There are also standard methods
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for integration, but for now
we're going to concentrate on
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the standard forms of
integration. The first one we're
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going to have a look at is the
integral of X to the N with
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respect to X.
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To do this? We take X to the
power N plus one, so we've added
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one to the index and then we
divide by the new index.
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We must remember to add on a
constant of integration, see.
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N can be any number,
whole number or fraction. A
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decimal except. It cannot be
minus one. If it were minus one,
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we have minus one, plus one
would be 0 and we're not allowed
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to divide by zero. So in this
particular case an does not
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equal minus one.
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An extension
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to this. Would
be a X plus B where
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again A&B are just numbers, just
constants again raised to the
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power N. AX plus B
is very little different to X,
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so again we had one to the index
and divide by the new index. But
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now because we're multiplying
our variable X by a, we've got
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to divide by it, because in
essence we are reversing the
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process of Differentiation.
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And again we must have source of
constant of integration, see and
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again we cannot have N being
equal to negative one.
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Next standard form is the
integral of one over X. Saw
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this when we did Differentiation
as we saw that the derivative
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of the log of X was
one over X. So if we
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reverse that process the
integral of one over X must
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be the log of X.
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And we must include the constant
of integration, see.
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If we have one over
a X plus BTX.
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And again, A&B are constants.
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Then what we must have is one
over A the log of AX plus
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B. Again, we must include a
constant of integration.
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Let's just have a look at
an example of this. The integral
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of one over 2 - 3
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XDX. Here we can see that
the A is negative three, so
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that's going to be minus one
over three log of the modulus of
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2 - 3 X plus. Again, the
constant of integration, see.
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Next standard form the integral
of E to the X
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with respect to X.
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We know that E to the X is its
own derivative, so when we
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integrate it, we must get
exactly the same form E to the X
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again. What if we've
got the to the MXDX?
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Again. Thinking as integration
as reversing differentiation. We
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need one over ME to the MX
because if we were to
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differentiate E to the MX, we'd
get me to the MX and we want
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the ends to cancel out and again
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plus C. Let's just have a
look at an example of that. The
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integral of E to the four XDX.
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And that will be 1/4 E
to the 4X plus a constant
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of integration, see.
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Will move on now and look at the
trig functions and look at the
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standard integrals associated
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with those. So the integral
of Cos X DX.
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The derivative of sine X is
cause eggs, so if we reverse
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that, the integral of cars must
be signed X again plus a
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constant of integration.
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If we integrate cause
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NX DX. Well, we
know what happens when we
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differentiate cynex. We get
Ensign Annex. So when we do the
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integration, we're going to have
the sign an X. But we've got to
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have one over N sign NX and
again plus the constant of
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integration, see. What about
doing sign integral of
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sine X? Again, we know
that the derivative of causes
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minus sign. So if we integrate
sign reversing the
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Differentiation, it's got to be
minus Cos X Plus C. And
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similarly if we have sign NX,
DX, then it's got to be minus
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one over N cause NX plus C.
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What
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about?
10
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Integral of Tan
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X. The X.
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We know that the integral of
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Tan. We can change to be
the integral of sign over cause
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the X. Now when we look
we can see that the top is the
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derivative of the bottom within
a minus sign, 'cause the
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derivative of causes minus sign,
and so we have minus the natural
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log of cause of X.
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When my subtracting a log where
dividing by what's inside and if
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we divide by cosine then that's
the same as SEK and so we
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can show that this is the log
of set of X and again we
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must have plus a constant of
integration because these are
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indefinite integrals. There's
another one that we can
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take and that is sex
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squared XDX. When we did, the
differentiation of Tangent, we
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saw that when you
differentiate it tan you gots
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X squared and so if we reverse
that, the integral of sex
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squared must be Tan X plus a
constant of integration, see.
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If we've got sex
squared NXDX, and again,
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this must be one
over N 10 of
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NX plus C.
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Now we're going to look at some
rather more complicated ones,
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but they are worthwhile
remembering because they can
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save an awful lot of work if you
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can identify them. These are
connected with the inverse trig
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functions. So we have the
integral of one over the square
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root of 1 minus X squared with
respect to X.
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With the differentiation of the
inverse trig function, we solve
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that when we differentiate it
signed to the minus one of X, we
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got this and so if we reverse
that sign to the minus one of X
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must be the answer to the
integral because integration
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reverses differentiation.
Similarly, we have something
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that looks very much the same. A
squared minus X squared. The X
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instead of 1 minus X squared.
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Then the integral is
signed to the minus
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One X over a plus C.
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Let's have a look at some
example using this result
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because sometimes we have to
work a little bit hard at this
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side. Let's take a
straightforward one, square root
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of 4 minus X squared.
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Now we can identify the four
with the A squared, so it's
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fairly clear that a must be
equal to two, so we end up with
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sign to the minus one of X over
2 plus see the constant of
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integration. What if it doesn't
quite look like that? What if
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we've got something here in
front of the X squared?
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So we take the integral
of one over the square
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root of 4 - 9
X squared DX.
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Here we can identify the A as
being too, but we've got 9 here.
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We really do need that just to
be an X squared, so we're going
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to do is take the nine out
through the square root. And why
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is it comes through the square
root, we will have to square
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rooted, which is going to leave
it as a three, and so we're
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going to come to this the
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integral of. One over 3.
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One over the square root of.
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For over 9 minus
X squared with respect
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to X now.
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This does look like one we've
done before. We can keep the one
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over three. That's fine.
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And now we need sign to the
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minus one. Of X over a?
What is a this time? Will a
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IS2 over three because a squared
is 4 over 9, so that's X
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over 2 over 3 plus C.
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Dividing by a fraction, we know
how to do that. We convert the
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fraction and multiply, which
gives us sign to the minus one
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of three X over 2 plus C.
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So we see we can do these quite
easily by trying to get that
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coefficient of X to be one and
not the number that's actually
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there. The other hyperbolic
function that we met was 10 to
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the minus one.
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And the derivative of 10 to
the minus one was this
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function one over 1 plus X
squared. So if we want to
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integrate one over 1 plus X
squared, the answer must be 10
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to the minus one of X Plus C.
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In the same way as we had an A
squared in there, let's put one
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in. Now in this one.
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The X.
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We get this one standard result,
one over a 10 to the
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minus One X over a plus
a constant of integration, see.
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Let's take an example
one over 9 plus X
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squared DX.
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Nine and a square to the same.
So a must be equal to three, so
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that gives us one over 310 to
the minus One X over 3 plus C.
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Same questions we had before.
What if it's not?
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A1 there what if there's a
number in there? Well, again,
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let's have a look. Let's take
the integral of one over.
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25 + 16
X squared DX.
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What do we do?
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We try to get a one there,
which means we take the 16 out
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so we'll have one over 16
integral of one over 25 over 16
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plus X squared DX.
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The one over 16 can stay
as it is.
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This is now a squared, so a must
be 5 over 4. So we want one
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over a one over 5 over 4.
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10 to the minus one of
X over a so X over
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5 over 4 plus the constant
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of integration. Now we need to
tidy that off a little bit. One
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over 16 * 1 over 5 over 4
is 4 over 510 to the minus one
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of four X over 5 plus the
constant of integration. See and
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we can make a simplification
Here by dividing by 4, giving us
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120th, four and four cancel and
4 * 5 is 20.
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10 to the minus one
4X over 5 plus ever
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present constant of integration,
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see. Now those are all the
standard forms. Once you've seen
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them and use them there. Well
worth learning learning because
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you can recognize them and you
can use them straight away and
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you don't have to worry about
how to do certain forms of
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Integration 'cause they just
there, they just part of what
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you do all the time, but
learning them is very important.