Genius! Brilliant! Exactly what I was looking for, and much more comprehensive and understandable than most other videos/forum threads on this integral!
thanks for also putting the section # and problem #! it makes the problem easy to find, this of course assuming everyone is using the same textbook! thank you!
Found an another way to solve this question. Let x = sinycosy On differentiating both sides we get: dx = cos(2y)dy Now back to the question: ∫sqrt(1 - 4x^2)dx = ∫sqrt(1 - 4sin^2ycos^2y)•cos(2y)dy = ∫cos^2(2y)dy [After simplifying] = (1/2)∫1 + cos(4y)dy = (1/2)y + (1/8)(sin4y) = (1/2)y + (1/4)sin2y•cos2y ...(1) Now, we know that x = sinycosy Therefore, 2x = sin(2y) ...(2) After constructing a right-angled ∆, we get: cos(2y) = sqrt(1 - 4x^2) ...(3) And from equation (2), we get: y = (1/2)arcsin(2x) ...(4) Putting the values of sin(2y), cos(2y) and y from equations (2), (3) and (4) respectively to equation (1), we get: (1/2)y + (1/4)sin2y•cos2y = (1/4)arcsin(2x) + (1/2)x•sqrt(1-4x^2) Hence, ∫sqrt(1 - 4x^2)dx = (1/4)arcsin(2x) + (1/2)x•sqrt(1-4x^2) + C Thank you..
Genio! Es un método diferente del que me enseñaron, y definitivamente me quedo con tu enseñanza. Gracias! (Genius! It's a different method than the one I was taught, and I definitely stick with your teaching. Thanks!)
Hi Sir Im trying to evaluate the integral of 1/sqrt(a^2-x^2) using both cos and sin substitution. With x=sinu I get arcsin(x/a) and with cosu I get -arccos(x/a) but when i substitute values for x and a I get different results why is this? Am i doing something wrong?
I love how you go through each step so that it's all so easy to understand. Thank you, and keep up the good work!
Genius! Brilliant! Exactly what I was looking for, and much more comprehensive and understandable than most other videos/forum threads on this integral!
THX from Brazil !!!! love ur guides ! thx
Hola hermano . Soy mexicano y veo tus videos desde ya hace un tiempo gracias a ti pase mi materia de cálculos integral. Gracias mil gracias
Como te va amigo?
Nice video, watching from India❤❤
thanks for also putting the section # and problem #! it makes the problem easy to find, this of course assuming everyone is using the same textbook! thank you!
Hi sir I am watching your video in India. ❤❤ Thanks sir for help solve to query questions
these videos are honestly the best
Thank you!
Honestly You're literally saving my life for my cal 2 course :D Keep up the good work, much love from Canada!
√(1-4x²): x belongs to [-1/2, 1/2]，so when x = 1/2 sinθ than -1
Thanks. I love to watch your tutorials. Great job.
thank you !! This video is 6 years old but still very relevant in 2021.
Thank you so much! I'm doing physics in college and never really got integration (apart from the basics) and you've really helped!
Thank you so much for these videos! They are invaluable and you explain very well.
Gracias chino, sólo vine buscando la parte del sen (2x), pero aún así se agradece lo que haces sigue así
Thanks!
You are saving me from wringing my brain for answering this equation for my daughter. This math for me was 20 years ago
Inday Girlie your daughter should look up how to solve it herself
Exactly what I was looking for. Thank you from England.
You saved me man. You are awesome!
i love you so much
ty
Thanks. Just started this. Reading through.
Thank you teacher keep going
Found an another way to solve this question.
Let x = sinycosy
On differentiating both sides we get:
dx = cos(2y)dy
Now back to the question:
∫sqrt(1 - 4x^2)dx
= ∫sqrt(1 - 4sin^2ycos^2y)•cos(2y)dy
= ∫cos^2(2y)dy [After simplifying]
= (1/2)∫1 + cos(4y)dy
= (1/2)y + (1/8)(sin4y)
= (1/2)y + (1/4)sin2y•cos2y ...(1)
Now, we know that x = sinycosy
Therefore, 2x = sin(2y) ...(2)
After constructing a right-angled ∆, we get:
cos(2y) = sqrt(1 - 4x^2) ...(3)
And from equation (2), we get:
y = (1/2)arcsin(2x) ...(4)
Putting the values of sin(2y), cos(2y) and y from equations (2), (3) and (4) respectively to equation (1), we get:
(1/2)y + (1/4)sin2y•cos2y
= (1/4)arcsin(2x) + (1/2)x•sqrt(1-4x^2)
Hence,
∫sqrt(1 - 4x^2)dx
= (1/4)arcsin(2x) + (1/2)x•sqrt(1-4x^2) + C
Thank you..
You could also substitute u=2x (, because 4x^2=(2x)^2), so that you get 1/(2sqrt(1-u^2))du, which integrated is 1/2arcsinu+C=1/2arcsin2x+C
thanks for all the help i REALLY appreciate your videos!
你讲的超级棒，比外国人讲的好！给你个赞哦
+Fangyu Ju 謝謝~
Thank you goat
Thank you for having such useful videos!!
Thx from DR 🇩🇴🇩🇴
Genio! Es un método diferente del que me enseñaron, y definitivamente me quedo con tu enseñanza. Gracias! (Genius! It's a different method than the one I was taught, and I definitely stick with your teaching. Thanks!)
thank you ^_^ ...... from Algeria
YOU ARE MY NEW GOD
thank you so much, your explanation was really helpful
Hi Sir Im trying to evaluate the integral of 1/sqrt(a^2-x^2) using both cos and sin substitution. With x=sinu I get arcsin(x/a) and with cosu I get -arccos(x/a) but when i substitute values for x and a I get different results why is this? Am i doing something wrong?
thanks!!!
As usual, I don't understand jack, but randomly in a year, I will magically understand everything. Very weird.
Thanks .. brilliant and easy to understand
What if "a" is not a perfect square? Will that matter in the integration?
You are a wonderful person
Thank you very much, it helped a lot to solve this problem
does it work for cosine substitution instead of sine?
Thanks from Zimbabwe
What's in your hand.i saw it first time.. amazing
Just a mic.
Its a pokeball
is it wrong to apply the formula directly?
Sir, I have the same question but with a + sign instead of a - sign. How do i do it please? Thank u
Use tan(x)
and if it was the Integral of sqrt(5-4x^2) or sqrt(3+4x^2)?
я наверное первый словянский коментатор , но что бы я без тебя делал >
google translate: "I'm probably the first Slavic commentator, but that I do without you"?
The correct translation has a little different meaning (like this): but what would I do without you
4:44 the integral of cosx is - sin not positive sin
Wrong
That is wrong the derivative of -sinx is -cosx NOT cosx
What if it's sqrt(4-y^2) is it the same process.. instead of x it's y
Hey sir, where did you study? your awesome btw :) im studying for A level math :)
I teach here in Los Angeles
Sos grande mostro!!
Thank u so much. it was so helpful
Why sqrt(cos^2)=cos? Why not |cos|?
Why cant we use x = cos u why do we have to use x = sin u?
JESUS FUCKING CHRIST THANK YOU SO MUCH IM UR BIGGEST FAN
Thanks from senegal :D
I'm late, but amazing explainer, thank you very much!
That final final final answer!!!
If on the place of "a" with out how it can solve
Sou brasileiro e não sei o que é mais difícil integral ou aprender inglês ks
It would be better if someone actually showed the triangle for this question...?
muito bom somente o resultado que deu errado e 1/4 arcoseno(2x) + 2x raiz de 1-4x^2
very good thanks a lot
Ok so sin(2∆) = 2sin∆cos∆,
What about cos(2∆)?
What about tan(2∆)?
Anyone?? I haven't been thought this anywhere
thankyou 😿
What if you ended with cos(2theta) instead of sin(2theta). Cos(2theta) has 3 identities..
Procel91 they'll all work
este momento donde la clase de ingles surte efecto
just let 2x = sinx 😢😢
Great.
nice!
that is a nifty mic
Good
Hii 2024❤
口音像华人!! where are you from?
melodynamics hi, 我是來自台灣
教学好好，谢谢你！I'm from malaysia!!
melodynamics 謝謝收看我的視頻。
I love you.
I love you too!!!
idol
Hypotenuse
Very hard
W
All that manipulation just to figure out the anti-derivative of sqrt. 1 - 4x^2. I'm sorry, but I hate Calculus lol.