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We already know how to handle some integration problems where the integrand

is a powers of sines and cosines. For example, I can anti-differentiate

sine to and odd power. How?

By trading in all but one of those sins for cosines.

Let's make this really concrete. Instead of talking about an odd power

well let's just make it 17. Can I antidifferentiate sin to the 17th

power? Yeah, I can rewrite this problem as sin

of x times sin squared x to the 8th power, because sin square to the 8th

power gives me 16 copies of sin times more sin gives me 17 copies of sin, now

we can trade sin squared for some cos signs right, sin squared is 1 minus cos

sign squared to the 8th power. So if I can do this anti-differentiation

problem all I need to do is do this anti-differentiation problem.

And I can do that with a substitution. So make the substitution u equals cosine

x and in that case d u is minus sign x dx.

Now maybe you're complaining you don't see a minus sign.

And you'll only see sin, but I'll just put a pair of cancelling minus sins

there, and I see a minus sin next dx so we got a du and the rest I can write in

terms of u. Specifically, this becomes negative, the

n is derivative of 1 minus u squared to the 8 power du, is what's left over and

this anti-differentiation problem that I can do and it is a little bit annoying

cause I guess I gotta expand this thing out to find an anti-derivative of this,

but I can do it. But what if instead I'd had an even power

of sine. Well, maybe I'll anti-differentiate sine

to the fouth power. All right?

So this is not an odd power but an even power.

that's harder, I can't trade in all of my sins for cosines because I need just one

left over to make the substitution work. Instead, I'll use an identity, the half

angle formula. It let's me replace sined squared of x by

1 minus cosine of 2x over 2. And it lets me replace cosign squared of

x by 1 plus cosign 2x over 2. How do those help?

Well I can rewrite this integral as the integral of sin squared, squared.

And now I can use this half angle identity.

This is the same as the integral of what sin squared.

Sin squared is 1 minus cos 2x over 2. All right?

These are equal. And still have to square that, dx.

Now, I expand. So, I get this is the same as

integrating. Well, the cos 2x over 2 term, squared is

cosigned squared of 2x over 4. And there's our cross term.

The 1 half times minus cosign 2 x over 2. And there's 2 of those.

So I end up getting minus 1 half. cosigned 2x.

And there's the one half term squared just plus a quarter, so I just have to do

this integration problem. I can split that into 3 integrals.

So this gives me the integral of cos sign squared 2x over 4dx minus the integral of

cos sign 2x over 2dx plus the integral of 1 4th dx.

Now, the second and third integration problem I can do, I'm just going to copy

down this one again. The integral of cos squared 2x over 4 dx.

This, here, well I could do this by making a substitution.

u equals 2x, say. And then I'll get that this is sine of 2x

divided by 4. And if I integrate a quarter, I get 1 4th

x. What about that first one?

How do I integrate cosine squared? So, to handle this, I can repeat the

trick with the half angle identity. But I'm looking at integrating cosign

squared 2x. So I'll replace x by 2x and I'll turn

this into a 4x. So I can use this identity.

How does that go? Well I get instead of this the integral

of 1 plus cosine 4x and instead of over 4 its now over 8 and then I'll just copy

down these things again. Sin 2x over 4 plus a quarter x.

Now I can put it all together. So I want to and differentiate 1 8th.

I get 1 8th x. And then I want to, and I differentiate

cos 4x over 8, and I get sin of 4x over 32 and then I'll include the rest.

So, I'll subtract sin of 2x over 4. I'll add a quarter x and I'll add some

constant. And I could combine the one quarter x and

the 1 8x, so I could right this as 3 8x plus sine of 4x over 32 minus sine of 2x

over 4 plus c. I should say that in some cases you can

get away with doing a bit less work. So I want to calculate the integral from

zero to pi of sine to the fourth and I'll again start the same way, right I'm going

to integrate from zero to pi and I'll write this as sine squared squared.

Just so I can see how the half angle formula is going to help me.

Alright, now I'll use the half angle formula as the integral from zero to pi

of 1 minus cosine2x over 2 squared, but now what do I want to do?

Well, I'll expand that out again, so this is the integral from 0 to pi of 1 fourth

Minus the cross term is one half cosign 2x and then plus the cosign times squared

2x over 4. And now there's a little trick.

I'm integrating from 0 to pi cosign. Right I'm integrating cosign over an

entire period. And that ends up being 0.

So I can just throw this whole term away and now I can keep on going.

I can also use the half angle formula here.

So this is the integral from 0 to pi. well I've still got the 1/4 plus, and

then what does this become? By the half angle formula, this is 1

plus. And instead of 2x.

It's cosine 4x over 8, but I would again, if I integrate cosine 4x, x going from 0

to pi, that's integrating cosine ove 2 complete periods that ends up cancelling.

So I can just throw that term away, and all I'm really integrating now is a

quarter plus an 8. Well thats 3 8ths, but I'm integrating

over an interval of life pi, so this definite integral is 3 8ths pi.

This turn out to be not so bad. See I'm getting an answer of 3 8ths pi,

but that's not really the point right? The cool thing about setting this up as a

definite integral is really just how easy it is to do the calculation since I can

throw away some terms along the way that I know would integrate to zero.