Could use the trig identity that has sin(x)+cos(x) = sqrt(2)*sin(x+pi/2) at the end to simply the result a little bit. It would be only in terms of sines at least.
Начиная с этого момента 4:25 перед членом sin(3u^2+6) стоит знак "-" дальше вы вводите обозначение I2, потом после решения. А с момента 9:17 знак потерян "+" вместо "-". Соответственно в решении 9:50 также пропущен знак "-".
Yes but unfortunately I got the colour palette wrong cuz the blue light filter here is alot different than the one I use on my phone.....I need to use brighter and softer colours and hopefully the next video is much better. Any suggestions on the colours? On the bright side yeah the extra screen real estate makes it easier to keep track of everything that's going on and my back and shoulders were quite comfortable while recording
@@maths_505Dude, wym? The colors looked just fine, they felt natural and didn't really bother me much. You already use some quite vibrant colors in your writing, so making the screen more vibrant/brighter would have those colors singe some eyeballs off. You could, if you want, modify the color pallete a tiny bit, but don't go anything further than a couple shades due to the aforementioned reason.
He's back! Finally! the last remaining days of my summer vacation have been bejeweled by the funny math man. I though of solving it by immediatly using the weistrauss sub, and then applying the trig identity to finish it off with IBP.
Glad you're back! One question, if the integral was - inf to +inf, what would the bounds of integration be? How would someone go about that? (I know in this case it's an even integrand but what if it wasn't?)
@@daddy_myers I mean if you do want to do a substitution like u=1/x, how would you modify the bounds of integration? Or you simply cannot because it's not a continuous function between the bounds? Are there rules for this? Does the function have to be continuous and maybe even monotic?
@@amidhmi5243 Generally, it's up to how nice the bounds are. As far as I'm aware, there's no general rule; it's more of an "eyeballing" thing where you just get it from experience. However, in this case, it doesn't matter whether or not you use that substitution, because an odd integrand integrated over a symmetric interval from -inf to +inf would yield zero. So, indeed, this transformation "works" because it reduces the integral to zero - but that's only because the integrand is odd; if it were neither odd nor even, then this transformation would be generally incorrect.
@@amidhmi5243 monotonicity and smoothness are sufficient to perform a substitution. If sub-function isn't monotonic, interval of integration has to be split into intervals of monotonicity, and substitution is performed separately for every integral over subinterval
god i missed you a lot dude
me too😊
The man is back with some more exciting maths. Let us all cheer!
I took a break for a while ....now I'm back and you'll see videos for the 2nd channel around the 7th or 8th of September
can you share the link of the second channel
@@maths_505
Please, how do you demonstrate that the integral from 0 to infinite of cos (x*2) equals square root of (Pi/8)?
Thanks a lot.
Could use the trig identity that has sin(x)+cos(x) = sqrt(2)*sin(x+pi/2) at the end to simply the result a little bit. It would be only in terms of sines at least.
Should be 'sin(x + π/4)'
@@GiornoYoshikage yeah you right my mistake
Начиная с этого момента 4:25 перед членом sin(3u^2+6) стоит знак "-" дальше вы вводите обозначение I2, потом после решения. А с момента 9:17 знак потерян "+" вместо "-". Соответственно в решении 9:50 также пропущен знак "-".
He's back!!! HE'S BACCCKKKK!!!!!
Sick new tablet, dude. I'm loving the new cozy & spacious view; it feels alot less cluttered.
Yes but unfortunately I got the colour palette wrong cuz the blue light filter here is alot different than the one I use on my phone.....I need to use brighter and softer colours and hopefully the next video is much better. Any suggestions on the colours?
On the bright side yeah the extra screen real estate makes it easier to keep track of everything that's going on and my back and shoulders were quite comfortable while recording
@@maths_505Dude, wym?
The colors looked just fine, they felt natural and didn't really bother me much.
You already use some quite vibrant colors in your writing, so making the screen more vibrant/brighter would have those colors singe some eyeballs off.
You could, if you want, modify the color pallete a tiny bit, but don't go anything further than a couple shades due to the aforementioned reason.
Okay bro
Check out the colour palette on the latest Instagram post. I think that one should work.
Looks perfect.
It's not too bright or too dark, and the shades look nice.
You have missed a minus before I 2 . Thank you for the exercise
Yes. So the last two terms of the final answer should both be negated.
Really it is interesting integral. Thank you for your smart solution.
He's back! Finally! the last remaining days of my summer vacation have been bejeweled by the funny math man. I though of solving it by immediatly using the weistrauss sub, and then applying the trig identity to finish it off with IBP.
Welcome back man, great video as always
√pi/8(3sin(2+pi/4)-(1/√3)*sin(6+pi/4)) is an easier way to writhe answer.
Bro which program do you use to write the solution?
Excellent work
The answer is more complicated than the original integral.
Amazing!
could you please suggest me textbooks for advanced calculus that demonstrate these techniques to solve these improper integrals?
Glad you're back!
One question, if the integral was - inf to +inf, what would the bounds of integration be? How would someone go about that? (I know in this case it's an even integrand but what if it wasn't?)
If it isn't an even integral, then the bounds of integration stay as they are and can only be modified under a transformation (substitution).
@@daddy_myers I mean if you do want to do a substitution like u=1/x, how would you modify the bounds of integration? Or you simply cannot because it's not a continuous function between the bounds? Are there rules for this? Does the function have to be continuous and maybe even monotic?
@@amidhmi5243 Generally, it's up to how nice the bounds are.
As far as I'm aware, there's no general rule; it's more of an "eyeballing" thing where you just get it from experience.
However, in this case, it doesn't matter whether or not you use that substitution, because an odd integrand integrated over a symmetric interval from -inf to +inf would yield zero.
So, indeed, this transformation "works" because it reduces the integral to zero - but that's only because the integrand is odd; if it were neither odd nor even, then this transformation would be generally incorrect.
@@amidhmi5243 monotonicity and smoothness are sufficient to perform a substitution. If sub-function isn't monotonic, interval of integration has to be split into intervals of monotonicity, and substitution is performed separately for every integral over subinterval
Nice bro
Hi math 505 from where do you get these integrals? Do you make them yourselves and verify with wolfram alpha?
Alot of times yes
ممتاز 😮😮😮😮😮😮
Nice, how the sqrt(pi/8) and √3 came out
Link in the description
Too much difficult. I lose heart when trying to solve integrals like this one.