Since I've finished uni exams, I've been pretty bored already to the point where I started making worked examples and notes for one of my uni/eng maths units from last semester. Anyway I thought some of you who may also be bored, and enjoy maths

, may be interested. If people are interested then I may make a few more like this, with some of the topics we covered (e.g. gauss elimination and such). So yeh, have a read through and try the questions if you want, and post the sols up if you can get them

.

I probably should point out that this is after spesh, and you won't be allowed to use these techniques in VCE maths. Although UMEP students would have probably covered this already.

Integration by PartsWe are all familar with the product rule, which we can rearrange to get the rule for integration by parts.

So we have two 'functions',

and

. The goal is to end up with a simpler integral than the original. This may be done by picking

as the right function, so that both

and the term

is anti-differentiable.

e.g.

So here, if we pick

and

, we will end up with the term

which is anti-differentiable, and a simpler integral than the first.

So basically its what you would have come across in Methods, integration by recognition, but backwards. So we have two 'functions',

and

. The goal is to end up with a simpler integral than the original. This may be done by picking

as the right function, so that both

and the term

is anti-differentiable.

e.g.

So here, if we pick

and

, we will end up with the term

which is anti-differentiable, and a simpler integral than the first.

We can also note that if we had of picked

and

the other way round, we would get

which is a worse integral than we started with.

So lets try another one.

e.g.

Now the little trick to this one is to look at it as

, that is

,

We may even need to apply integration by parts twice, as sometimes the first integral won't be any simpler (note: this won't always work).

You don't have to show all that working, but I've kept it all there for now.

One last example, sometimes when we use integration by parts twice, we can obtain the same integral, as the one we were originally looking for, in the RHS, and add it back to the other side, then solve for it to leave what the integral equals.

e.g.

As we can see we know have the original integral

on the RHS as well, so we can 'solve' for it by adding it to the left hand side and dividing by 2.

Now for some questions to try (answers are in the spolier tags).

We may also sometimes use euler's identity

to use a simpler method to find the anti-derivative of some of the above. But first we are going to show one way of why

. Firstly we start off with a differential equation,

with the initial condition of

. So we are looking for a function that basically is it's own derivative, so we can use

. Now what happens if we let

?

We can check that this satisfies

,

.

Now we know that

will be a solution to the differential equation, but if we look at another function,

we can see that it is a solution as well.

Now checking with the initial condition,

Since both solutions are solutions ot the same equation, and satisfy the initial condition, we can say that they are the same.

So now that we have seen one way of showing euler's identity, we can use it to find some anti-derivatives.

Lets start off looking at

We can find the anti-derivative of s similar function involving

, and then take the real part of it (we use the real part for

and the imaginary part for

).

So to find the real parts we remember that a real part multiplied by a real part gives a real number, and the imaginary part multipled by an imaginary part gives a real part (as

). So that is the

multipled by the

and the

multiplied by the

.

Which gives the same result as the other method, you wouldn't have to show all these steps, which means sometimes this method is quicker.

If we were looking at a function with

instead of

then we would be looking for the imaginary part instead of the real part, which means that when we get to the multiplying out line, we will be looking for the real parts multiplied by imaginary parts to give an imaginary part.

Now for some questions to try.