www.mathcentre.ac.uk/.../9.3%20Use%20of%20a%20table%20and%20standard%20rules.mp4

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.