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Trig integration by linear transformation

JPencil

On 'the hard integral thread' I have been seeing integration of trig expressions done by transformations such as

I was wondering if anyone had any resources about this topic?

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Reply 1

Louisb19

Original post by JPencil

On 'the hard integral thread' I have been seeing integration of trig expressions done by transformations such as

I was wondering if anyone had any resources about this topic?

It's something I used to during preparation for MAT.

It's not really a transform, you just have to be able to recognise things like

\sin{(x)} = \sin{(x +2k\pi)} = \sin{(\pi - x)} = sin{(\pi (2k - 1) - x)}

Similarly

\cos{(x)} = \cos{(-x)}

ect...

Just consider the graphical representations of these transformations, you should see what we are getting at with this.

If this is not what you are referring to forgive me.

Reply 2

TeeEm

for any integral with limits a to b and integrand f(x)

replacing f(x) with f(a+b-x), without touching the limits, leaves the integral invariant,
(easily proved by substitution)

Reply 3

JPencil

Original post by Louisb19

It's something I used to during preparation for MAT.

It's not really a transform, you just have to be able to recognise things like

\sin{(x)} = \sin{(x +2k\pi)} = \sin{(\pi - x)} = sin{(\pi (2k - 1) - x)}

Similarly

\cos{(x)} = \cos{(-x)}

ect...

Just consider the graphical representations of these transformations, you should see what we are getting at with this.

If this is not what you are referring to forgive me.

That is what I am referring to. How would these kinds of transforms (i don't know what else to call them) change the limits of the integral?

Reply 4

JPencil

Original post by TeeEm

for any integral with limits a to b and integrand f(x)

replacing f(x) with f(a+b-x), without touching the limits, leaves the integral invariant,
(easily proved by substitution)

Thank you, i'll prove it for myself

Reply 5

TeeEm

Original post by JPencil

Thank you, i'll prove it for myself

what you are referring in your first post is just a special case

Reply 6

JPencil

Original post by TeeEm

what you are referring in your first post is just a special case

Thank you very much. PRSOM

Reply 7

Louisb19

Original post by JPencil

That is what I am referring to. How would these kinds of transforms (i don't know what else to call them) change the limits of the integral?

It depends. If you have the following integral

\displaystyle\int_b^a \sin{(x)} \mathrm{d} x

We can show that this is equal to

\displaystyle\int_b^a \sin{(\pi - x)} \mathrm{d} x

Using the identity (the linear transform of sin(x)) from my first post. You are not performing a integral transform, you are just using

\sin{(x)} = \sin{(\pi-x)}

. The limits will not be changed.

If we now consider

\displaystyle\int_b^a \cos{(\pi -x)} \sin{(x)} \mathrm{d} x

Under

x \rightarrow \pi - x

we have

\displaystyle\int_{\pi - b}^{\pi - a} \cos{(x)} \sin{(\pi - x)} \mathrm{d} x = \displaystyle\int_{\pi - b}^{\pi - a} \cos{(x)} \sin({x}) \mathrm{d} x

You can see that since we are performing an integral transform here the limits will be changed. It's a little hard for me to explain since we are talking about two types of transformations; Linear and Integral (I think this is what it is called). Hopefully this still helps though.

(edited 8 years ago)

Reply 8

JPencil

Original post by Louisb19

It depends. If you have the following integral

\displaystyle\int_b^a \sin{(x)} \mathrm{d} x

We can show that this is equal to

\displaystyle\int_b^a \sin{(\pi - x)} \mathrm{d} x

Using the identity (the linear transform of sin(x)) from my first post. You are not performing a integral transform, you are just using

\sin{(x)} = \sin{(\pi-x)}

. The limits will not be changed.

If we now consider

\displaystyle\int_b^a \cos{(\pi -x)} \sin{(x)} \mathrm{d} x

Under

x \rightarrow \pi - x

we have

\displaystyle\int_{\pi - b}^{\pi - a} \cos{(x)} \sin{(\pi - x)} \mathrm{d} x = \displaystyle\int_{\pi - b}^{\pi - a} \cos{(x)} \sin({x}) \mathrm{d} x

It does help, a lot actually. Thank you for being so patient. When will i know to change the limits? and what to change them to?

Reply 9

username1162972

Original post by JPencil

It does help, a lot actually. Thank you for being so patient. When will i know to change the limits? and what to change them to?

If the linits are 0 to pi and its a function of sinx use pi/2-x as this changes it to cos and limits to -pi/2 to pi/2 then since cosx is even remove the -pi/2 for 0 and multiply by two.
Say some integrand is 0 to infinity id use 1/x sub the limits are the main clue.

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Reply 10

JPencil

Original post by physicsmaths

Is there anywhere I can read about this topic? What is the specific name? Thanks by the way

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Reply 11

username1162972

Original post by JPencil

Is there anywhere I can read about this topic? What is the specific name? Thanks by the way

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I don't think so, it just comes with practice tbh just practice STEP and it will come naturally!

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Reply 12

JPencil

Original post by physicsmaths

I don't think so, it just comes with practice tbh just practice STEP and it will come naturally!

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Which STEP paper are they mostly on? I'm doing STEP I & II

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Reply 13

username1162972

Original post by JPencil

Which STEP paper are they mostly on? I'm doing STEP I & II

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Yes it will come up on them. Start from year 2000

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Reply 14

TeeEm

Original post by JPencil

Is there anywhere I can read about this topic? What is the specific name? Thanks by the way

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one of the best affordable books on integration which i highly recommend is

INTEGRAL CALCULUS
by Dhami

it retails new between £7 and £15

And I want to say this
The best mathematics books are "Made in India"
I own around 12 and each and every one of them I have read/used cover to cover
If I was to go there I would probably buy hundreds

Reply 15

Gregorius

Original post by JPencil

Is there anywhere I can read about this topic? What is the specific name? Thanks by the way
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This is quite an awkward question to answer as the topic generally seems to be classified as one of those "tricks" that you learn about for doing integrals - so there's little systematic material to be found. One possible source is the recent book "Inside Interesting Integrals". If you go to the web page indicated, you'll find a pdf extract of chapter 2 that includes section 2.2 devoted to examples of this technique.

Perhaps the trouble is that the problem is too general and too specific at the same time. If you want to evaluate

\displaystyle I = \int_a^b f(x) dx

The the substitution

u = (a+b) - x

will give you

\displaystyle I = \int_a^b f(a+b-u) du

What you really want to happen then is something like

\displaystyle f(a+b-u) = g_1(a+b) + g_2(a+b) f(u)

which then allows you to pull out

\displaystyle I = \frac{(b-a)g_1(a+b)}{1-g_2(a+b)}

So to use the method, you are dependent upon the addition properties of the integrand; I've illustrated one functional relation that will work, but there would appear to be many other possibilities...and this is just for the very simple linear transform

u = (a+b) - x

Reply 16

JPencil

Original post by Gregorius

\displaystyle I = \int_a^b f(x) dx

The the substitution

u = (a+b) - x

will give you

\displaystyle I = \int_a^b f(a+b-u) du

What you really want to happen then is something like

\displaystyle f(a+b-u) = g_1(a+b) + g_2(a+b) f(u)

which then allows you to pull out

\displaystyle I = \frac{(b-a)g_1(a+b)}{1-g_2(a+b)}

u = (a+b) - x

I was actually looking at 'inside interesting integrals' a few weeks ago, I must have forgotten :mmm:

. I see what you mean about this being more of a 'trick' than a method. Thank you for the reply, it's very helpful, this will be essential for STEP!

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Reply 17

JPencil

Original post by TeeEm

I will defiantly invest in some books of that kind, thank you. Will this book be helpful for Alevels/ STEP or mainly for undergrad? I'm trying to prioritise my reading to match my syllabus :smile:

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Reply 18

TeeEm

Original post by JPencil

I will defiantly invest in some books of that kind, thank you. Will this book be helpful for Alevels/ STEP or mainly for undergrad? I'm trying to prioritise my reading to match my syllabus :smile:

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caters for integration for every level

Reply 19

atsruser

Original post by Gregorius

So to use the method, you are dependent upon the addition properties of the integrand

I think that it's worth mentioning that one of the most frequently useful properties is that of periodic symmetry, which crops up mainly with trig functions.

So we use the fact, say, that

\int_0^{\pi/2} f(\sin x) \ dx = \int_{\pi/2}^{\pi} f(\sin x) \ dx

and generalisations of that, or that

\tan(\frac{\pi}{2} - x) = \cot x = \frac{1}{\tan x}

and so on.

To prove the first result algebraically, you can do the following:

Let

I = \int_{\pi/2}^{\pi} f(\sin x) \ dx

Now use the fact that

\sin(\pi -x) = \sin \pi \cos x - \cos \pi \sin x = \sin x

by creating a shifted set of (u,y)-axes where

u=\pi-x \Rightarrow x=\pi-u

.

We have

x=\frac{\pi}{2} \Rightarrow u=\frac{\pi}{2}

and

x=\pi \Rightarrow u=0

.

We also have, by hand-waving, that

\frac{dx}{du} = -1 \Rightarrow dx=-1 du

Now we can say that

I = \int_{\pi/2}^{\pi} f(\sin x) \ dx = \int^{u=0}_{u=\pi/2} f(\sin (\frac{\pi}{2}-u)) \cdot {-1} \ du

This says that the area measured by the first integral on the graph with (x,y)-axes is the same as that measured by second integral on the graph with the shifted (u,y)-axes. But there are couple of things to note now:

1. The use of

u

is irrelevant - it's a dummy variable used to name our new axes, and we can, without fear of confusion, replace it with

x

. So now:

I = \int^{x=0}_{x=\pi/2} f(\sin (\frac{\pi}{2}-x)) \cdot {-1} \ dx

If this confuses you, imagine that you've drawn two sets of axes on a page, one labelled with x and one with the shifted (u,y)-axes. Once you start to look purely at the (u-y)-axes, you won't get confused if you replace

u

with

x

, and sometimes people write this algebraically by

x \to \pi-x

and change the integral similarly - but that's hard to do at first (or if you're thick, like me).

2. The lower limit is bigger than the upper limit. That's weird. However, remember that an integral is an area computed from thin strips parallel to the y-axis. In the normal case, we imagine thin strips moving in the positive direction, from say 0 to 1, and in this case, we can think of as

dx

being a small positive vector, pointing along the width of each thin strip.

Here, however, we have

{-1} \ dx

- that's a small *negative* vector, pointing along the width of each thin strip - if we add up a load of small negative vectors, we get a big negative vector i.e. we expect the integral to cover an area from, say, 1 to 0, or in this case, from

\pi/2

to 0.

But, clearly, in both cases we calculate the same area - so we can a) change

{-1} \ dx

dx

and b) flip the limits, while getting the same result. So now:

I = \int_{x=0}^{x=\pi/2} f(\sin (\frac{\pi}{2}-x)) \ dx

.

3. And finally, from the trig result above, we can write:

I = \int_{x=0}^{x=\pi/2} f(\sin x) \ dx

as required.

That is a lot of work, of course. In real life, you can do the same thing much quicker by staring a graph of trig functions, and noting where the symmetries occur, and then writing down the transformed integral directly from the picture - you can always prove it formally as above if necessary.