
Bulgarian: 
Кривата в тъмносин цвят е
графиката на функцията
r = 1 – cos(θ),
като работим с полярни координати.
Интересно ми е дали
можем да намерим
площта, оградена от тази крива.
Насърчавам те да спреш
видеото и да опиташ самостоятелно.
Сега да го направим заедно.
Вече видяхме каква
е логиката на формулата,
че площ, оградена от
крива в полярни координати,
ще бъде равна на 1/2 от
определения интеграл
от началния ъгъл тита
до крайния ъгъл тита,
от алфа до бета, от (r(θ))^2).dθ.
Сега просто трябва да я приложим
към тази функция ето тук.

English: 
- So this darker curve in blue
is the graph of r is equal
to 1 minus cosine of theta,
of course we're dealing
in polar coordinates here.
And what I'm interested in is
to see if we can figure out
the area enclosed by this curve.
And I encourage you to pause the video
and try it on your own.
Alright, let's work through it together.
So we've already seen, we've
already given ourselves
the intuition for the formula,
that the area enclosed by a polar graph
is going to be equal to one
half the definite integral
from our starting theta
to our ending theta,
from alpha to beta, of r
of theta squared d theta.
And so we, essentially,
just have to apply this
to this function right over here.

Portuguese: 
Esta curva mais escura em azul
é o gráfico de r é igual
a um menos cosseno de teta,
claro que estamos lidando
com coordenadas polares.
E o que eu estou interessado é
em ver se podemos descobrir
a área delimitada por essa curva.
E eu o encorajo a pausar o vídeo
e tentar por conta própria.
Tudo bem, vamos trabalhar juntos.
Nós já vimos, já nos foi dada
a intuição para a fórmula,
que a área delimitada
por um gráfico polar
será igual a metade
da integral definida
do nosso teta inicial ao
nosso teta final,
de alfa até beta, de r
de teta ao quadrado d teta.
E assim nós, essencialmente,
temos que aplicar isto
nesta função bem aqui.

Thai: 
เส้นโค้งสีฟ้าเข้ม
คือกราฟของ r เท่่ากับ 1 ลบโคไซน์ของเธต้า
แน่นอน เราคิดพิกัดเชิงขั้วตรงนี้
และสิ่งที่ผมสนใจ คือดูว่าเราหา
พื้นที่ที่ล้อมรอบด้วยเส้นโค้งนี้ได้ไหม
ผมแนะนำให้คุณหยุดวิดีโอ
แล้วลองทำเองก่อน
เอาล่ะ ลองทำไปด้วยกัน
เราเห็นแล้ว เราได้สัญชาตญาณ
สำหรับสูตร
พื้นที่ที่ล้อมรอบด้วยกราฟเชิงขั้ว
ว่าเท่ากับ 1/2 อินทิกรัลจำกัดเขต
จากเธต้าเริ่มต้นถึงเธต้าสุดท้าย
จากอัลฟาถึงเบต้า ของ r กำลังสองเธต้า d เธต้า
แล้ว เราก็แค่ต้องใช้สูตรนี้
กับฟังก์ชันนี่ตรงนี้

Korean: 
그래서 이 파란색의 어두운 곡선은
r=1-cosθ의 그래프입니다
물론 우리는 여기서
극좌표를 다루고 있습니다
그리고 제가 하고 싶은 것은
이 곡선으로 둘러싸인 영역의 넓이를
알아낼 수 있냐는 것입니다
한번 동영상을 멈추고
혼자 시도해보길 바랍니다
이제 같이 해봅시다
우리는 이미 직감적으로 공식을 떠올렸습니다
우리는 이미 직감적으로 공식을 떠올렸습니다
극 그래프에 의해 둘러싸인 영역의 넓이는
시작점의 θ에서 끝점의 θ까지
½∫(r(θ))²dθ과 같습니다
시작점의 θ에서 끝점의 θ까지
½∫(r(θ))²dθ과 같습니다
여기서는 시작점의 θ를 α로
끝점의 θ를 β로 하겠습니다
그래서 우리는 이 상황을
함수에 적용시키기만 하면 됩니다

English: 
So in this case, the area
is going to be equal to
one half the definite integral.
Now what's our alpha and what's our beta?
Well, we're going from theta
is equal to zero radians,
and we're essentially going all the way--
When theta is equal to
zero radians is 1 minus 1
we're right over there.
And then we go all the way around
to theta is equal to two pi radians.
Notice when we're back at two
pi, cosine of two pi is one,
one minus one is zero again.
So we get back to that point.
So we're going from theta
is equal to zero radians
to theta is equal to two pi radians.
Now what's r of theta squared?
Maybe I'll color code this a little bit.
R of theta squared.
Well it's just going to be
one minus cosine of theta.
One minus cosine theta squared.
And of course we have our d theta.
We have our d theta.
And now we just have to
evaluate this integral.
So once again, at any
point you feel inspired,

Korean: 
이 경우에는 영역이
정적분한 것의 절반과 같습니다
그럼 우리의 α와 β는 무엇일 것 같습니까?
우리는 θ가 0라디안일 때 시작해서
이렇게 쭉 따라갈 겁니다
θ가 0라디안일 때 1-1이고
바로 여기에 있을 때입니다
그리고 θ가 2π라디안이 될 때까지
쭉 돕니다
2π라디안에 다시 도달하면
cos2π는 1이기 때문에
1-1은 다시 0이 됩니다
결국 다시 같은 지점으로 온 것입니다
그래서 시작점 θ는 0라디안이고
끝점 θ는 2π라디안입니다
그럼 (r(θ))²는 무엇일 것 같습니까?
(r(θ))² 부분을 이 색으로 표현하겠습니다
(r(θ))² 부분을 이 색으로 표현하겠습니다
그냥 1-cosθ입니다
(1-cosθ)²이고
물론 뒤에 dθ도 있습니다
물론 뒤에 dθ도 있습니다
이제 적분만 하면 됩니다
여러분이 원한다면 언제든지 스스로

Portuguese: 
Então neste caso, a área
será igual a
metade da integral definida.
Agora, qual é o nosso alfa
e qual é o nosso beta?
Bem, nós estamos indo de teta
é igual a zero radianos,
e estamos indo até o fim--
Quando teta é igual a zero
radianos é um menos um
estamos bem aqui.
E então, percorrermos
todo o caminho
até teta é igual a dois pi radianos.
Observe que quando voltamos a dois
pi, cosseno de dois pi é um,
um menos um é zero novamente.
Por isso, voltamos a esse ponto.
Estamos indo de teta
é igual a zero radianos
até teta é igual a dois pi radianos.
Agora o que é r de
teta ao quadrado?
Eu vou colorir um pouco.
r de teta ao quadrado.
Bem, isso será um
menos cosseno de teta.
Um menos cosseno de
teta ao quadrado.
E claro, temos o nosso d teta.
Temos o nosso d teta.
E agora só temos de
avaliar essa integral.
Mais uma vez, quando
se sentir inspirado,

Bulgarian: 
В този случай площта
ще бъде равна на
1/2 от определения интеграл.
Кои са тук алфа и бета?
Започваме от θ = 0 радиана,
и изминаваме целия път до...
Когато тита е равна на 0 радиан,
това е 1 – 1,
тогава сме тук.
И после се придвижваме до
тита е равно на 2π радиана.
Обърни внимание, че когато сме
при 2π, косинус от 2π е 1,
1 – 1 е равно на нула отново.
Връщаме се в тази точка.
Значи се движим от
θ = 0 радиани
до θ = 2π радиана.
Колко е r(θ)^2?
Може би трябва да
използвам различни цветове.
r(θ)^2
Това е равно на (1 – cos(θ))^2.
И, разбира се, имаме dθ.
Сега само трябва да
изчислим този интеграл.
Пак напомням – във всеки момент,
когато почувстваш вдъхновение,

Thai: 
ในกรณีนี้ พื้นที่จะเท่ากับ
1/2 อินทิกรัลจำกัดเขต
ตอนนี้ อัลฟาคืออะไร และเบต้าคืออะไร?
เราจะไปจากเธต้าเท่ากับ 0 เรเดียน
และเราจะไปจนถึง --
เมื่อเธต้าเท่ากับ 0 เรเดียนคือ 1 ลบ 1
เราอยู่ตรงนี้
แล้วเราวนรอบ
ไปถึงเธต้าเท่ากับ 2 พายเรเดียน
สังเกตว่าเมื่อเรากลับมาที่ 2 พาย
โคไซน์ของ 2 พายเป็น 1
1 ลบ 1 เป็น 0 เหมือนเดิม
เรากลับมาที่จุดนั้น
เราจะไปจากเธต้าเท่ากับ 0 เรเดียน
ถึงเธต้าเท่ากับ 2 พายเรเดียน
แล้ว r ของเธต้ากำลังสองเป็นเท่าใด?
บางที ผมจะใช้สีแทนความหมายหน่อย
r ของเธต้ากำลังสอง
มันจะเท่ากับ 1 ลบโคไซน์ของเธต้า
1 ลบโคไซน์เธต้ากำลังสอง
และแน่นอน เรามี d เธต้าของเรา
เรามี d เธต้า
และตอนนี้เราต้องหาค่าอินทิกรัลนี้
เหมือนเดิม ถ้าคุณรู้สึกอยากลองตอนไหน

Portuguese: 
tente calcular isso.
Então, vamos fazer isto.
O que eu faria ...
Isto será igual a um meio
vezes a integral definida
de zero até dois pi.
Deixe-me expandir isso.
Isto será um menos
dois cosseno teta
mais cosseno ao quadrado teta, d teta.
d teta.
Sei como tirar a
antiderivada de um,
Sei a antiderivada de
cosseno negativo de teta,
mas cosseno de teta ao quadrado,
isto é um pouco ...
Ele parece algo
que você possa fazer,
use substituição U ou algo parecido,
mas por sorte, temos nossas
identidades trigonométricas,
e por isso sabemos.
Sabemos que cosseno ao
quadrado de teta
é exatamente a mesma coisa
que um meio vezes um
mais cosseno de dois teta.
Aprendeu isto na
aula de trigonometria.
Se não aprendeu, você
aprenderá agora.
E é por isso que esta é
uma das mais úteis

Korean: 
적분해보길 바랍니다
이제 같이 해봅시다
제가 할 것은
0에서 2π까지 적분하고
그것의 절반을 구하는 것입니다
이것을 전개해보겠습니다
전개하면 (1-2cosθ+cos²θ)dθ가 됩니다
전개하면 (1-2cosθ+cos²θ)dθ가 됩니다
전개하면 (1-2cosθ+cos²θ)dθ가 됩니다
우리는 1을 어떻게 부정적분하는지 알고
-cosθ도 부정적분할 수 있지만
cos²θ는 약간 어렵습니다
cos²θ는 약간 어렵습니다
보자마자 바로 알긴 어렵고
치환적분이나 비슷한 것을 해야합니다
운이 좋게도, 우리에겐
삼감함수 반각 공식이 있고
우리는 cos²θ이
우리는 cos²θ이
½(1+cos2θ)와 같다는 것을
알고 있습니다
삼각법 시간에 배웠었습니다.
아직 안 봤다면, 지금 배웠으니 괜찮습니다
삼각함수 공식들 중에서도

Thai: 
ก็ลองหาค่านี้ได้
ลองทำดูกัน เอาล่ะ
สิ่งที่ผมจะทำ --
อันนี้จะเท่ากับ 1/2
คูณอินทิกรัลจำกัดเขตจาก 0 ถึง 2 พาย
ขอผมกระจายอันนี้ออกนะ
อันนี้จะเท่ากับ 1 ลบ 2 โคไซน์เธต้า
บวกโคไซน์กำลังสองเธต้า d เธต้า
d เธต้า
ตอนนี้เรารู้วิธีหาปฏิยานุพันธ์ของ 1
เรารู้วิธีหาปฏิยานุพันธ์ของ
ลบโคไซน์ของเธต้า แต่โคไซน์กำลังสองเธต้า
อันนี้ค่อนข้าง --
คุณอาจไม่เห็นทันทีว่าคุณทำได้
ใช้การแทนที่ u หรืออะไรพวกนั้น
แต่โชคดีของเรา เรามีเอกลักษณ์ตรีโกณมิติ
เราจึงรู้
เรารู้ว่าโคไซน์กำลังสองของเธต้า
จะเท่ากับ 1/2 คูณ 1
บวกโคไซน์ของ 2 เธต้า
คุณเรียนไปในวิชาตรีโกณมิติ
ถ้าคุณไม่ได้เรียน คุณก็ได้เรียนแล้วตอนนี้
และนั่นคือสาเหตุที่เอกลักษณ์นี้มีประโยชน์

Bulgarian: 
опитай да решиш това 
самостоятелно.
Сега да го направим заедно.
Аз бих направил следното...
Значи това е равно на 1/2 по 
определен интеграл от 0 до 2π.
Сега ще го разложа.
Това е равно на
1 – 2cos(θ)
плюс (cos(θ))^2, dθ.
Знаем как да намерим
примитивната функция на 1,
знаем как да намерим
примитивната функция на
минус косинус от тита,
но квадрата на косинус тита,
това е малко сложно...
Не ти хрумва веднага
как може да стане,
да използваме интегриране
чрез заместване или нещо такова,
но за наш късмет имаме
тригонометричните тъждества.
Знаем, че квадратът на
косинус тита
е равен на 1/2 (1 + cos(2θ)).
Това се учи в тригонометрията.
Ако не сте го учили,
го научаваш в този момент.
Това е едно от най-полезните

English: 
try to evaluate this.
So let's do this, alright.
So what I would do...
So this is going to be equal to one half
times the definite integral
from zero to two pi.
And let me expand this out.
This is going to be one
minus two cosine theta
plus cosine squared theta, d theta.
D theta.
Now I know how to take the
anti-derivative of one,
I know how to take the anti-derivitive of
negative cosine of theta,
but cosine squared theta,
this is a little bit...
It doesn't jump out at you
that you can just do this,
use u-substitution or something like that,
but lucky for us, we have
our trigonometric identities,
and so we know.
We know that cosine squared of theta
is just the same thing
as one half times one
plus cosine of two theta.
You learned this in trigonometry class.
If you didn't, we'll,
you've learned it just now.
And that's why this is
one of the more useful

Bulgarian: 
тригонометрични тъждества,
когато трябва да намериш
примитивната функция на нещо
или когато интегрираш.
Да го направим.
Да препишем това тук като
1/2 (1 + cos(2θ)).
Да видим, може би ще можем...
Ще го направя ето така.
Предполагам, че ако искаме...
Добре, ще го направя така.
Това е равно на 1/2
и после ще...
1/2, нека да започнем с
определянето на примитивната функция.
1/2.
Примитивната функция 
на 1 спрямо тита
е равна на тита.
Примитивната функция на
–2cos(θ) е...
това е равно на –2sin(θ).
Можеш да намериш производната,
производната на синус
е косинус,

English: 
trigonometric identities if
you're finding any type of
anti-derivative or if
you're integrating anything.
And so let's do that.
Let's rewrite this right over here as
one half times one plus
cosine of two theta.
And let's see, and maybe we could...
Yeah, let's just do it like that.
I guess we could, if we wanted...
Well, we'll just do it like that.
So this is going to be equal to one half,
and then we are going to...
One half, now let's just
start taking anti-derivatives.
One half.
Now the anti-derivative of
one, with respect to theta,
is just going to be theta.
The anti-derivative of
negative two cosine of theta,
well that's just going to be
negative two sine of theta.
Negative two sine theta.
You can take the derivative,
the derivative of sine is cosine
and the negative two,
it'll just multiple it

Korean: 
여러분이 무언가를 부정적분하거나 적분할 때
반각 공식이 특히 유용하다는 것을 알 수 있습니다
그럼 시작해봅시다
여기있는 cos²θ를
½(1+cos2θ)로 바꿔씁시다
그리고 천천히 살펴보면
방법이 보일 겁니다
제 생각에는
이렇게 하는 게 좋을 것 같습니다
일단 ½을 쓰고
일단 ½을 쓰고
부정적분을 시작해봅시다
부정적분을 시작해봅시다
1을 θ에 대해 부정적분하면
θ가 될 것이고
-2cosθ를 부정적분하면
-2sinθ가 될 것입니다
-2sinθ가 될 것입니다
여러분은 미분을 할 수 있습니다
sin을 미분하면 cos이고
-2는 sinθ를 미분한 것에 곱해집니다

Portuguese: 
identidades trigonométricas caso
você encontre qualquer tipo de
antiderivada ou se está
integrando alguma coisa.
Vamos fazer isto.
Vamos reescrever
isto aqui como
um meio vezes um mais
cosseno de dois teta.
Vamos ver, talvez pudéssemos ...
Sim, vamos fazer assim.
Acho que poderíamos,
se quiséssemos ...
Bem, vamos fazer assim.
Então, isto será igual a meio,
e depois vamos ...
1/2 - agora vamos apenas
tomar as antiderivadas.
1/2.
Agora, a antiderivada de
um, em relação à teta,
será apenas teta.
A antiderivada de
dois cosseno de teta negativo,
isto será apenas dois
seno de teta negativo.
Dois seno de teta negativo.
Tomando a derivada,
a derivada de seno é cosseno
e o dois negativo,
apenas irá multiplicar

Thai: 
ถ้าคุณหาปฏิยานุพันธ์
อะไรก็ตาม หรือถ้าคุณจะอินทิเกรตอะไรก็ตาม
ลองทำกันดู
ลองเขียนอันนี้ตรงนี้เป็น
1/2 คูณ 1 บวกโคไซน์ของ 2 เธต้า
แล้วลองดู บางทีเราอาจ --
ใช่ ลองทำแบบนั้นดู
ผมว่าเราทำได้ ถ้าเราต้องการ --
เราจะทำแบบนั้น
อันนี้จะเท่ากับ 1/2
แล้วเราจะ --
1/2 ตอนนี้เราจะหาปฏิยานุพันธ์
1/2
ทีนี้ ปฏิยานุพันธ์ของ 1 เทียบกับเธต้า
จะเท่ากับเธต้า
ปฏิยานุพันธ์ของลบ 2 โคไซน์ของเธต้า
มันจะเท่ากับลบ 2 ไซน์ของเธต้า
ลบ 2 ไซน์เธต้า
คุณหาอนุพันธ์ได้
อนุพันธ์ของไซน์คือโคไซน์
และลบ 2 มันจะคูณ

Portuguese: 
a derivada de seno de teta
temos dois cosseno negativo de teta.
E então teremos, vejamos ...
Deixe-me distribuir isto.
Isto é a mesma coisa
que um meio mais meio de
cosseno de dois teta.
Vamos supor desta forma.
Assim, a antiderivada de um meio,
a antiderivada de um meio.
Estou olhando para aquilo ali.
Será um meio teta.
Um meio teta.
E então a antiderivada de um meio
cosseno de dois teta, vamos ver ...
A derivada de seno de dois teta
é dois cosseno de dois teta.
Assim, a antiderivada disto é ...
A antiderivada de
cosseno de dois teta,
pode fazer substituição U se quiser,
mas você é capaz de
fazer isto em sua cabeça,
a antiderivada de cosseno
de dois teta será
meio de seno de dois teta.
E então você tem este
meio bem aqui.
Então, isso será ...

Bulgarian: 
и –2, просто ще ги умножа
по производната на sin(θ),
значи –2cos(θ).
После получаваме...
Ще разкрия скобите.
Това е равно на 1/2 + 1/2cos(2θ).
Да го приемем по този начин.
Примитивната функция на 1/2.
Сега ще разгледам това тук.
Това е (1/2)θ.
Примитивната функция на
(1/2)cos(2θ), да видим...
Производната на sin(2θ)
е 2cos(2θ).
Значи примитивната функция
на това е...
Примитивната функция на
косинус от 2θ,
тук можеш да интегрираш
със заместване, ако искаш,
но може да успееш да го
направиш и наум,
примитивната функция
на косинус от 2θ
е 1/2 синус от 2θ.
И после остана това 1/2 тук.
Това ще бъде...

Thai: 
กับอนุพันธ์ของไซน์ของเธต้า
มันก็คือลบ 2 โคไซน์ของเธต้า
แล้วเราจะได้ ลองดู --
ขอผมแจกแจงอันนี้นะ
อันนี้เท่ากับ 1/2 บวก 1/2
โคไซน์ของ 2 เธต้า
ลองคิดมันแบบนี้
ปฏิยานุพันธ์ของ 1/2
ปฏิยานุพันธ์ของ 1/2
ผมกำลังดูมันตรงนี้
มันจะเท่ากับ 1/2 เธต้า
1/2 เธต้า
แล้วปฏิยานุพันธ์ของ 1/2
โคไซน์ของ 2 เธต้า ลองดู --
อนุพันธ์ของไซน์ของ 2 เธต้า
คือ 2 โคไซน์ของ 2 เธต้า
แล้วปฏิยานุพันธ์ของตัวนี้คือ --
ปฏิยานุพันธ์ของโคไซน์ของ 2 เธต้า
และคุณแทนที่ u ได้ถ้าต้องการ
แต่คุณอาจจะอยากคิดในใจ
ปฏิยานุพันธ์ของโคไซน์ของ 2 เธต้าจะเป็น
1/2 ไซน์ของ 2 เธต้า
แล้วคุณมี 1/2 นี่ตรงนี้
ค่านี้จึงเท่ากับ --

Korean: 
-2는 sinθ를 미분한 것에 곱해집니다
그래서 -2cosθ입니다
그 다음으로
½(1+cos2θ)에 분배법칙을 적용합시다
그러면 ½+½cos2θ가 됩니다
그러면 ½+½cos2θ가 됩니다
이 상태로 계속 진행합시다
½을 부정적분하면
½을 부정적분하면
이 부분이
½θ가 됩니다
½θ가 됩니다
이제 ½(cos2θ)를 부정적분 해봅시다
이제 ½(cos2θ)를 부정적분 해봅시다
sin2θ를 미분한 값은
2cos2θ와 같으므로
이것을 부정적분한 것은
cos2θ를 부정적분한 것은
원한다면 치환적분도 할 수 있지만
이것은 암산으로도 할 수 있습니다
cos2θ를 부정적분 한 것은
½sin2θ와 같습니다
그리고 여기에 ½이 있습니다
어떻게 할지 고민해봅시다

English: 
times the derivative of sine of theta
so it's negative two cosine of theta.
And then we're going to have, let's see...
So let me distribute this.
This is the same thing
as one half plus one half
cosine of two theta.
So let's just assume it's this way.
So the anti-derivative of one half,
so the anti-derivative of one-half.
So I'm really looking at
that right over there.
It's going to be one half theta.
One half theta.
And then the anti-derivative of one half
cosine of two theta, let's see...
The derivative of sine of two theta
is two cosine of two theta.
So the anti-derivative of this is...
The anti-derivative of
cosine of two theta,
and you can do u-substitution if you like,
but you might be able
to do this in your head,
the anti-derivative of cosine
of two theta is going to be
one half sine of two theta.
And then you have this
one half right over here.
So this is going to be...

Korean: 
제가 무엇의 부정적분을
하고 있는지 보여주겠습니다
바로 이것과
이것이 되겠습니다
그래서 ¼sin2θ를 여기에 더하면 됩니다
만약 여기 마지막 부분이 헷갈린다면
이것의 미분을 하기를 권장합니다
sin2θ를 미분한 것은
2cos2θ이고
¼에 2를 곱하면 ½이므로
½cos2θ를 얻는 것입니다
그리고 2π와 0을 대입합시다
그리고 2π와 0을 대입합시다
그래서 답을 구할 때 아마 한 가지가
눈에 금방 띌 것인데 0을 대입할 때입니다
이 식 전체가 모든 항이 0이 될 것이고
그래서 모든 것이 단순화될 것입니다
그래서 2π의 경우만 구해서
반으로 나누면 됩니다
그래서 2π의 경우만 구해서
반으로 나누면 됩니다
이것은 ½에 2π를 곱한 것과 같고
sin2π는 0입니다
그러니까 그냥 0이 됩니다
½에 2π를 곱한 것은

English: 
Let me show you what I'm
finding the anti-derivative of.
Of that right over there, of this,
and I guess this right over here.
So this is going to be plus
one fourth sine of two theta.
And I encourage you to
find the derivative here
if that last part was
a little bit confusing.
The derivative of sine of two theta
is two cosine of two theta,
two over one fourth is one half,
you get to one half cosine of two theta.
And we're going to
evaluate that at two pi,
at two pi, and at zero.
So when you evaluate
it, one thing that might
jump out at you is when
you evaluate this at zero,
this whole thing,
everything, every term here
is just going to be zero, so
that simplifies things nicely.
So we really just have
to take one half of...
It evaluated at two pi.
So this is going to be
one half times two pi,
two pi, and then sine of two pi is zero,
so that's just going to be zero.
And then plus one half times two pi,

Thai: 
ขอผมแสดงให้ดูว่าผมหาปฏิยานุพันธ์ได้อะไร
ของตัวนั้น ของตัวนี้
ตรงนี้
อันนี้จะเท่ากับบวก 1/4 ไซน์ของ 2 เธต้า
และผมแนะนำให้คุณลองหาอนุพันธ์ตรงนี้ดู
ถ้าส่วนสุดท้ายนั้นทำให้คุณงง
อนุพันธ์ของไซน์ของ 2 เธต้า
คือ 2 โคไซน์ของ 2 เธต้า
2 ส่วน 1/4 คือ 1/2
คุณจะได้ 1/2 โคไซน์ของ 2 เธต้า
และเราจะหาค่ามันที่ 2 พาย
ที่ 2 พายกับที่ 0
เมื่อคุณหาค่ามัน สิ่งหนึ่งที่
คุณอาจเห็นคือว่า เมื่อคุณหาค่านี้ที่ 0
ทั้งหมดนี้ ทุกเทอมตรงนี้
จะเท่ากับ 0 มันจึงลดรูปสวยงาม
เราแค่ต้องหา 1/2 ของ --
หาค่าที่ 2 พาย
อันนี้จะเท่ากับ 1/2 คูณ 2 พาย
2 พาย แล้วไซน์ของ 2 พายเป็น 0
มันก็แค่ 0
แล้วบวก 1/2 คูณ 2 พาย

Bulgarian: 
Нека да ти покажа как намирам 
примитивната функция
на това ето тук, от това,
и това ето тук.
Значи това ще стане
+1/4 синус от 2θ.
Препоръчвам ти да намериш
производната,
ако последното те обърква.
Производната на синус от 2θ
е 2 по косинус от 2θ.
2 върху 1/4 е равно на 1/2.
Получаваме 1/2 косинус от 2θ.
И сега ще го сметнем
за 2π и за 0.
Докато го смяташ, може да ти
хрумне, че
когато го смяташ за нула,
цялото това нещо тук,
всеки член тук
просто ще бъде нула,
което много добре се опростява.
Просто имаме 1/2 от това,
сметнато за 2π.
Това е равно на 1/2 по 2π,
после синус от 2π е нула,
така че това тук е нула.

Portuguese: 
Deixe-me mostrar do que estou
tomando a antiderivada.
Disto aqui,
e acho que disto aqui.
Isto será mais um quarto
seno de dois teta.
E eu o encorajo a encontrar
a derivada aqui
se esta última parte foi
um pouco confusa.
A derivada de seno
de dois teta
é dois cosseno de dois teta,
dois sobre quarto é um meio,
você obtém meio de
cosseno de dois teta.
E estamos calculando
isto em dois pi,
em dois pi, e em zero.
Então, quando você calcular,
uma coisa que pode
perceber é que quando
calcula isto em zero,
esta coisa toda,
tudo, cada termo aqui
será apenas zero, e
isto simplifica bem as coisas.
Temos apenas que calcular
um meio de...
É calculado em dois pi.
Portanto, isto será
um meio de dois pi,
dois pi, e seno de
dois pi é zero,
então isto será zero.
E depois mais um
meio vezes dois pi,

Portuguese: 
isto será mais pi.
E então duas vezes
dois pi, seno de quatro pi,
que ainda será zero.
Isto será zero também,
e estamos quase terminando.
Portanto, isto será
um meio vezes três pi
ou três meios, três
meios pi é a área,
é a área desta região.
Legendado por [Raul Guimaraes].
Revisado por [Pilar Dib]

English: 
so that's going to be plus pi.
And then sign of two times
two pi sine of four pi,
that's still going to be zero.
So this is going to be zero as well,
and we are almost done.
So this is going to be
one half times three pi
or three halves, three
halves pi is the area,
is the area of this region.

Thai: 
มันจึงเท่ากับบวกพาย
แล้วไซน์ของ 2 คูณ 2 พาย ไซน์ของ 4 พาย
ยังจะเท่ากับ 0
อันนี้จึงเท่ากับ 0 เช่นกัน
เราใกล้เสร็จแล้ว
อันนี้จะเท่ากับ 1/2 คูณ 3 พาย
หรือ 3/2, 3/2 พายคือพ้นที่
พื้นที่ของเขตนี้

Bulgarian: 
След това 1/2 по 2π
става плюс π.
После синус от 2π по 2
е синус от 4π,
това също е нула.
Значи и това е нула,
и сме почти готови.
Това е равно на 1/2 по 3π,
или 3/2 по π е площта
на тази област тук,
оградена от графиката
на функцията.

Korean: 
π이므로 더해줍시다
sin(2×2π)는 sin4π이므로
여전히 0입니다
결국 이 또한 0이 되고
이제 거의 끝났습니다
½×3π을 정리하면 3π/2입니다
½×3π을 정리하면 3π/2입니다
이 값이 이 영역의 넓이입니다
커넥트 번역 봉사단 | 류한준
