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