Thank you, Asian Ball Man. You've not only helped me remember a missing fragment of the nightmare that was calc II, but you've also taught me to love. To love with the power of the ball.
Sin3x=3sinx - 4sin^3x From this formula , Sin^3x=3sinx-sin3x/4 Now we can easily integrate by taking 1/4 outside and giving integral sign to both of them ...🙂
Yes you are completely right but no one ever remembers this formula (at least i never remembered it) though they know about it.What we saw in the video was more "practical" nevertheless whatever people are comfortable with should be used.
[yet another "thank-you" comment] Got here after not understanding the steps my teacher followed to reach the final answer. Awesome video and awesome, super clear explanation. Thanks a lot :)
Helpful, I start to forget the most basic trig the higher I go... was trying to use the sin^2x=1/2(1-cos2x) for some reason and thats why I couldn't get it...
I always like videos like this. Furthermore, the video made the process easy to understand. I know this would be a slightly different way of going about solving the integral, but would it not be simpler, using u-substitution, to manipulate du into equaling sinxdx? This would result in -du, which could be plugged back into the integral, then the negative of du (-1 constant) could be brought to the front of the integral.
That is lovely. Thank you I feel so daft, 60 years too late to find it all making senseMight as well make this my bucket list... learn another integral....might even try to do some myself before looking to you for the answer.
+blackpenredpen I mean use that when doing the cos^2(x)*sin(x). Your link is the exact rule again, but you did a substitution again. The substitution does the same thing, but is just an extra unnecessary step as this is a reverse chain rule.
yeah it was really great but is the thing in your hand for translating to english or for recording sound . but i really enjoyed the video so much even subscribed already and also hit the 👍👍👍👍👍👍👍👍👍👍👍👍👍✔
Error occurs when carrying sin over and cancelling sin x. He should have substituted the sinxdx with -du instead. Although the answer is correct but the way is incorrect which makes will lead to wrong understanding of derivatives.
I just learnt complex numbers and cams back to look at this vid. Ah how beautiful to see the times where I struggled with integration of weird trigo functions, now with complex numbers it’s almost like cheating.
mesmerised by your pen switching sleight of hand
This comment made my day!!!
+blackpenredpen i strongly sympathise with this comment. and with your sleight of hand. pens for the win!
+Daniel Walther Thank you, thank you~
btw, there's a video on that th-cam.com/video/akYITCn0Ecg/w-d-xo.html
Oooo i see...... ❤
I appreciate you and your microphone immensely :)
david112755 lol. Thank you
Who's watching this in 2024
Alternative:
Use : sin3x=3sinx-4sin³x
sin³x=(3sinx-sin3x)/4
∫ sin³x dx = ∫ (3sinx-sin3x)/4
on integrating = -(3cosx/4)+cos3x/12
👍
But The Answers Are Completely Different How??
@@raghav9o9 Substitute cos3x =4cos³x - 3cosx. Will be getting the same answer.
So i am confused can't we use formula of
Cos2x=1-2sin²x
sin²x =1-cos2x/2
This is exactly what I was done
In the U world
All my classes are online now because of the coronavirus, but these videos are saving me. Thank you bprp 🙏
I like how this is 3:14
+Dan Dart 3,14159265358979
+Dan Dart edit:3.1415926535897932384626432795028841971
+a pool player that's wrong after 4626, it goes 46264338327950288419716939937510
Taking it up another step 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899
3.141592653589793238462643383279502884197169399375105820974944592307816406286 208998628034825342117067982148086513282306647093844609550582231725359408128481 117450284102701938521105559644622948954930381964428810975665933446128475648233 786783165271201909145648566923460348610454326648213393607260249141273724587006 606315588174881520920962829254091715364367892590360011330530548820466521384146 951941511609433057270365759591953092186117381932611793105118548074462379962749 567351885752724891227938183011949129833673362440656643086021394946395224737190 702179860943702770539217176293176752384674818467669405132000568127145263560827 785771342757789609173637178721468440901224953430146549585371050792279689258923 542019956112129021960864034418159813629774771309960518707211349999998372978049 951059731732816096318595024459455346908302642522308253344685035261931188171010 003137838752886587533208381420617177669147303598253490428755468731159562863882 353787593751957781857780532171226806613001927876611195909216420198938095257201 065485863278865936153381827968230301952035301852968995773622599413891249721775 283479131515574857242454150695950829533116861727855889075098381754637464939319 25506040
Explanation was crystal clear.....
The way switch the colour of the marker is just cool
I love you teaching method ,me from India
Awesome video. But why are you talking into your shower head?
diamondsmasher ㅋㅋㅋlaughing out loud
lol
Thank you, Asian Ball Man. You've not only helped me remember a missing fragment of the nightmare that was calc II, but you've also taught me to love. To love with the power of the ball.
I love how this video's length is the value of π
you're brilliant
I'm actually finishing my bachelor engineering, sometimes we forget, but with your video I was back on track, thanks a lot !
i like math... seemed more interesting than the rest showing up on youtube.
this is what i call entertainment while being ill!
Math is fun!
+blackpenredpen yup
Wis I felt the same way :(
@@blackpenredpen ya
Sin3x=3sinx - 4sin^3x
From this formula ,
Sin^3x=3sinx-sin3x/4
Now we can easily integrate by taking 1/4 outside and giving integral sign to both of them ...🙂
Yes
Yes you are completely right but no one ever remembers this formula (at least i never remembered it) though they know about it.What we saw in the video was more "practical" nevertheless whatever people are comfortable with should be used.
So i am confused can't we use formula of
Cos2x=1-2sin²x
sin²x =1-cos2x/2
sir , you teach so smooth and so clear .
Ur talking expressions are totally different from us...but instead of this...it's so helpful..🇮🇳🇮🇳🇮🇳
thank you mate you saved my night
Omg. That's the first integral from your channel I solved by myself. Your videos are excellent!
You are really amazing sir. jazakallahu kair. May lord guide u on straight path
Your voice is जबरदस्त... खुप छान...
You used substitution method nicely.no matter what are you using it matters how are you solving.very great substitution and 5 🌟 for adjusting. By
[yet another "thank-you" comment]
Got here after not understanding the steps my teacher followed to reach the final answer. Awesome video and awesome, super clear explanation. Thanks a lot :)
this one save me so Thank you. I was so clueless about the lesson our instructor send us but through this, now it make sense.
muchas gracias hermano , me suscribo de una . Explicas demasiado bien las integrales
Your explanation is super to dupper thank you sir . I purple you 💜💜🤩🤩
LEAN
LEAN
Sir when we use formula. sin³x = 3sinx- sin 3x ÷4 answer is -3÷4cosx+1÷12cos3x +c
Isn't it just sin^4(x)/4cos(x), using the backwards chainrule?
fx^(n+1)/(n+1)f'x
Sorry I'm new to this.
The pen switching was lit! Thanks so much!
Great explanation! I could never figure out how to where the negative sign came from. Thanks for the awesome break down!
these all really really help me to understand what's the correct way. thankiee!
Thankyou! It Helped me to cross check my answer
my o levels are in like few months, and i gotta say, this is actually really helpful bro, thanks for the sweet vid dude
You make it look easy. Thanks.
THANK YOU, i have a midterm tmr, and I did not understood substitution properly with trig. This was very helpful!
Thank you from 🇩🇿
Gracias por la explicación. Are you does videos about series of potential?
even after taking diff eq and getting past fourier transforms and stuff after that. i still forget things. thank you mate
شكرا صاحبي معلم
Helpful, I start to forget the most basic trig the higher I go... was trying to use the sin^2x=1/2(1-cos2x) for some reason and thats why I couldn't get it...
I am so grateful for this
Thank you, saved my homework.
Excellent solution
Superb sir explained superbly
Thanks for allowing me to pass calc mate.
I always like videos like this. Furthermore, the video made the process easy to understand. I know this would be a slightly different way of going about solving the integral, but would it not be simpler, using u-substitution, to manipulate du into equaling sinxdx? This would result in -du, which could be plugged back into the integral, then the negative of du (-1 constant) could be brought to the front of the integral.
try doing it by coverting it in triple angle. sin3x=3sinx-4sin^3x
It works nicely by parts
That is lovely. Thank you I feel so daft, 60 years too late to find it all making senseMight as well make this my bucket list... learn another integral....might even try to do some myself before looking to you for the answer.
wow that's great!!
I am from India ,really amazing sir
Thanks a lot i've bees searching for one hour this ^^
You look very different compared to now.
"U" nailed this nicely, and I am a SUB. (:o)
+lexinaut Thank "U" too!! =D
The video is π minutes . Perfection 😎👌
Superb tutorial. Thanks a lot man!
Nice video. It's also solvable without substitution (integration by parts):
∫ sin^3(x)dx =
∫ sin^2(x)sin(x)dx =
-cos(x)sin^2(x) + 2 ∫ sin(x) cos^2(x)dx
∫ sin(x) cos^2(x)dx =
-cos^3(x) - 2 ∫ sin(x) cos^2(x)dx
3∫ sin(x) cos^2(x)dx = -cos^3(x)
∫ sin(x) cos^2(x)dx = -1/3cos^3(x)
Plug this back into the original equation:
∫ sin^3(x)dx =
-cos(x)sin^2(x) - 2 * -1/3cos^3(x)=
-cos(x)sin^2(x) - 2/3cos^3(x)
So...
∫ sin^3(x)dx = -cos(x)sin^2(x) - 2/3cos^3(x)
Which simplifies to:
∫ sin^3(x)dx = -cos(x) + 1/3cos^3(x)
Aint no one got time for that
Sin3x=3sinx-sin^3x this formula can be used for integration....
comedy ka khazana sab ko hasana sin3x= 3sinx-4sin^3x
u r a bit wrong Gaurav
it should be sin^3(θ)={3sin(θ)−sin(3θ)}/4
Thanks, that was a great refresher.
Thank you for saving my life. AGAIN!
But the answer comes out to be -3/4 cosx + 1/12 cos3x + C . By using the identity sin3x = 3sinx - 4sin³x
wow you're amazing teacher thanks a lot
substitution not necessary, just use integral f'(x)*f(x)^n=(1/(n+1))*f(x)^(n+1)+c
+tifnatandmat but we didn't have cos(x)
check this out! th-cam.com/video/kWyzNjjrTBQ/w-d-xo.html
+blackpenredpen I mean use that when doing the cos^2(x)*sin(x). Your link is the exact rule again, but you did a substitution again. The substitution does the same thing, but is just an extra unnecessary step as this is a reverse chain rule.
+tifnatandmat Ah I see, you meant the U-substitution part. I thought you meant the trig identity part.
All good buddy! We're both right :P
+tifnatandmat yup yup!
Thank you!! I was desperate and you helped me a lot!
glad to help!
THANK YOU SO MUCH FOR THIS; V HELPFUL
You could also use the formula sin^3x=(3sinx- sin3x)/4
Very well explained, thank you
why u dont use the integral of power function,∫ sin³x dx=1/4(-cos x) sin^4(x) .why
Wonderfully explained!! 😊😊
How to solve if it has double integration and two limits that are 0 to π and 0 to 2π
Awesome explanation, thank you!
explained it better in 3 minutes than my AP AB Calc teacher did in 50
Thanks. This was really helpful!
who else was worried at -integral(1-u^2du) where it should be -integral((1-u^2)du)
seriously i spent 5 minutes trying to figure out how he just took out the parentheses!
Creative & Random also he forgot +constant after the integration, but then placed it after subbing cosx.
Meeeeeeee
yeah it was really great but is the thing in your hand for translating to english or for recording sound . but i really enjoyed the video so much even subscribed already and also hit the 👍👍👍👍👍👍👍👍👍👍👍👍👍✔
Danke dir fur es.
Fantastisch!
Thanks for solving doubt
You can skip a step by distributing the negative sign. Simply flip
∫ -(1-u²) du
like so:
∫ u²-1 du
duh
U did it great but I really liked ur baowl
omg this came up in my exam the other day
Great job! Thanks a lot! Easy and vivid!
I am glad to help!!!
Very helpful! Thank you
Fratè si gruoss!!!! (You're great bro! in Napoletan)
sir ihave opinion that we can directly use du instead of -sinx dx and take the negative sign out of the integral
Thank You. Very well explained. :)
So i am confused can't we use formula of
Cos2x=1-2sin²x
sin²x =1-cos2x/2
thanks me sirvió.. very good !!!
amazees me every time
Awesome video :) , suscribed!
Thank you very much. Teacher
Me dio gusto resolver este problema sin haber visto la solución.
Error occurs when carrying sin over and cancelling sin x. He should have substituted the sinxdx with -du instead. Although the answer is correct but the way is incorrect which makes will lead to wrong understanding of derivatives.
+mixeltastic I see what you mean, but the way I do it was sub dx = du/-sinx, which I do have that necessary neg in my steps.
Thank you for your help
Thank you sir
Very plain explanation. Good job :)
Sir what is the integral of ((sinx)^3)/x with limit 0 to infinity?
Isn't it: -( u - 1/3*U^3 * U) ?
Because the integral of dU is U so the result should be -(U-1/3*U^4).
Am I right ? ( don't talk too rich i'm french 😂)
Thank you alot sir 😊😊😊😊♥️♥️♥️♥️
Awesome explanation 🔥🔥🔥
I just learnt complex numbers and cams back to look at this vid. Ah how beautiful to see the times where I struggled with integration of weird trigo functions, now with complex numbers it’s almost like cheating.
Can you do a video on solving the indefinite integral, from 0 to infinity, sin(x^2) dx or sin^2(x) dx ?