Using trig identity to use u substitution Khan Academy English Czech

Using trig identity to use u substitutionYouTube↗ IA↗(long shot)
Education 127.9K views 86 likes 0 dislikes 23 Oct 2014 Show Contributors 1

Khan Academy 9.4M subscribers United States YouTube↗

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Thai Tee Ponsukcharoen

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- [Voiceover] Let's see if we can take the indefinite integral of cosine of X to the third power.

Podívejme se, zda dokážeme najít primitivní funkci k 'cos(x) na třetí'.

I encourage you to pause the video and see if you can figure this out on your own.

Vyzývám vás, pozastavte si video a zkuste to nejdříve sami.

You have given it a go and you might have gotten stuck.

Pokud jste to zkusili, mohli jste se někde zaseknout.

Some of you all might have been able to figure it out,

Někteří jste to vyřešili,

but some of you all might have gotten stuck.

někteří jste nevěděli, jak dál.

You're like, "Okay, cosine to the third power.

Řekli jste si: „Ok, 'cos(x) na třetí',

"Well, gee, if I only had a derivative of cosine here,

kež bych tu měl derivaci kosinu.

"if I had a negative sign of X or a sin of X here, "maybe I could've used U substitution,

Kdybych tu měl sin(x), mohl bych využít substituce.

"but how do I take the anti-derivative "of cosine of X to the third power?"

Jak ale najdu primitivní funkci k 'cos(x) na třetí'?“

The key here is, is to use some basic trigonometric identities.

Klíčem je využít některých základních trigonometrických identit.

What do I mean by that?

Co tím myslím?

We know that sin squared X plus cosine squared X

Víme, že 'sin(x) na druhou' plus 'cos(x) na druhou' je roven 1.

is equal to one, or if we subtract sin squared

Odečteme-li 'sin(x) na druhou',

from both sides, we know that cosine squared X is equal to one, write it this way, is equal to one minus sin squared X.

dostaneme, že 'cos(x) na druhou' je roven 1 minus 'sin(x) na druhou'.

What would happen if cosine to the third power, that's cosine squared times cosine.

Co když máme 'cos(x) na třetí'? To je cos(x) krát 'cos(x) na druhou'.

What happens if we were to take

Co když dosadíme za 'cos(x) na druhou'?

that cosine squared?

Let me just rewrite it. This is the same thing as cosine of X times cosine squared of X, DX.

Je to stejné jako cos(x) krát 'cos(x) na druhou' krát dx.

What if we were to take this thing

Co kdybychom vzali tuto věc…

right over here, let me do that magenta color.

Udělám to fialově.

What if we were to take this right over here and replace it with this.

Co kdybychom vzali tuto věc a nahradili ji tímto?

I now what you're thinking.

Vím, co si myslíte.

"Sal, what's that going to do for me? "This feels like I'm making this integral "even more convoluted."

„Sale, jak mi to pomůže? Děláš ten integrál ještě horší.“

What I would tell you, I would say,

Na to bych vám řekl:

"This might seem like it's getting more complicated,

„Zdá se, že to je komplikovanější,

"but as you explore and you play with it, "you'll see that this actually makes "the integral more solvable."

ale jakmile to trochu upravíme, uvidíte, že takto jej vyřešíme snadněji.“

Let's try it out.

If we do that, this is going to be equal to the indefinite integral, cosine of X times one minus sin squared X, DX.

Bude to tedy neurčitý integrál z cos(x) krát (1 minus 'sin(x) na druhou') krát dx.

What is this going to be equal to?

Čemu to bude rovno?

This is going to be equal to,

let me do this in that green color.

Udělám to zeleně.

This is going to be equal to the indefinite integral of cosine X.

Bude to rovno integrálu z cos(x)…

I'm just going to distribute the cosine of X.

Roznásobím to tím cos(x).

Cosine of X minus, minus cosine of X, cosign of X sin squared of X, sin squared, sin squared X and then I can close the parentheses, DX.

cos(x) minus [cos(x) krát 'sin(x) na druhou']. uzavřu závorku, krát dx.

This, of course, is going to be equal to the integral of cosine of X, DX, and we know

To bude samozřejmě integrál cos(x) dx, to víme, čemu se to bude rovnat,

what that's going to be, minus the integral.

minus integrál… Změním barvu.

I'll switch to one color now, of cosine of X, sin squared X, sin squared X, DX.

Integrál cos(x) krát 'sin(x) na druhou' krát dx.

Now, this is where it gets interesting.

To je velmi zajímavé.

This part right over here is pretty straight forward.

Tato část je snadná.

The anti-derivative of cosine of X is just sin of X.

Primitivní funkce k cos(x) je sin(x).

This right over here is going to be sin of X.

Toto bude sin(x).

I'll worry about the plus C at the end because both of these

O konstantu se budu zajímat až nakonec,

are going to have a plus C, so might as well just put one big plus c at the end.

oba integrály dají nějakou konstantu, takže je dám později dohromady.

That's sin of X and then what do we have going on over here?

Toto je tedy sin(x) a co máme tady?

Well, you might recognize, I have a function of sin of X.

Určitě poznáte, že mám funkci závislou na sin(x),

I'm taking sin of X and I'm squaring it

Mám sin(x) a mocním ji na druhou.

and then I have sin of X's derivative right over here.

Rovněž tu mám derivaci sin(x).

This fits the, I have some derivative of a function

Mám derivaci nějaké funkce, označím f'(x)

and then I have another and then I have a, I guess you could say, a function of that function. G of F of X.

a mám tu i funkci té funkce, označím g(f(x)).

That's a sin that maybe U substitution is in order,

To napovídá, že využijeme substituci.

or we've seen the pattern, we've seen this show

Tento vzor jsme viděli už několikrát.

multiple times already, that you could just say, "Okay, if I have a function of a function "and I have that functions derivative,

Můžete říct, že máte funkci nějaké funkce a také její derivaci,

"then essentially I can just take the anti-derivative "with respect to this function." This would be equal to, say, capital G is the anti-derivative of lower case G.

můžete integrovat vzhledem k té funkci. G je primitivní funkce k g.

Capital G of F of X plus C.

G(f(x)) plus C.

Now, if what I said didn't make sense, then we could do

Pokud vám nedávalo smysl to, co jsem právě řekl,

U substitution and go through it a little bit more step by step.

uděláme substituci a vezmeme to krok po kroku.

Let's just do that because we want things to make sense.

Udělejme to tak, chceme, aby to dávalo smysl.

That's the whole point of these videos.

To je celý smysl těchto videí.

We could say U is equal to sin of X and then DU is going to be equal to cosine of X, DX.

Označme 'u' rovno sin(x) a 'du' bude cos(x) dx.

This part and that part is going to be DU

Toto bude tedy 'du'

and then this is going to be U squared.

a toto bude 'u na druhou'.

This is going to be minus.

Toto bude minus.

We have the integral of U squared, DU.

Máme integrál 'u na druhou' krát du.

What is this going to be?

This is going to be, we're going to have negative U to the third power over three.

Bude to 'u na třetí' lomeno 3.

Then, we know what U is. The U is equal to sin of X.

Víme, co je 'u', 'u' je sin(x).

We have our sin of X here for the first part of the integral, for the first integral.

Z prvního integrálu máme sin(x).

We have the sin of X and then this is going to be minus.

Let me just write it this way.

Napíšu to takto.

Minus 1/3 minus 1/3.

(-1 lomeno 3) krát…

Instead of U to the third, we know U is sin of X.

Namísto 'u na třetí', víme, že 'u' je sin(x).

Sin of X to the third power.

'sin(x) na třetí'.

Then now, we can throw that plus C there.

Nakonec přičteme konstantu C.

We've just evaluated that indefinite integral.

Spočítali jsme neurčitý integrál.

The key to it is to just play around a little bit with trigonometric identities so that you can get

Klíčem k řešení bylo pohrát si s trigonometrickými identitami,

the integral to a point that you can use the reverse chain rule or you can use U substitution,

abychom jej dostali do tvaru, kdy umíme použít substituci.

which is just really another way of expressing the reverse chain rule.

Což je v podstatě zpětné řetězové pravidlo pro derivaci.

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