I did trig sub on the integral of vdu (x = tan(theta)) and got the answer, but the "add 1 and subtract 1" option is really cool. I need to remember that.
This is actually a way of connecting arctan with complex logarithms, which are defined in terms of log of magnitude and imaginary part of the angle of the complex number, so it's not necessarily a bad exercise to see what happens then!
Depends on the level of the audience. When I taught techniques of integration in high school none of my students were versed in complex numbers well enough. So IBP would have been the absolute go to for this question. Followed by a 3 sec long division.
That actually works. since Complex numbers have an Euler representation (z = |z| e^(i arg(z))) log(z) can be related to arg(z) and |z|. arg(z) links into arctan etc. On simplification it'll give the same thing
My thought was to integrate ln(x+i) + ln(x-i) dx -> (x+i)ln(x+i) - x - i + (x-i)ln(x-i) - x + i + C = xln(1+x^2) - 2x + i ln((x+i)/(x-i)) + C [ i ln((x+i)/(x-i)) = i ln((1-ix)/(-1-ix)) = 2iartanh(-ix) = -2i^2 arctan(x), using identity artanh(ix) = iarctan(x) ] = xln(1+x^2) - 2x + 2arctan(x) + C
It seems poly long division gets such a bad rap on TH-cam math videos. It takes just a couple of seconds and is so easy and reliable. Adding zero hmmm.
I did trig sub on the integral of vdu (x = tan(theta)) and got the answer, but the "add 1 and subtract 1" option is really cool. I need to remember that.
That was a nice warm up for the day. Thank you.
You are very wise my senior, thanks so much bro
The add 1 subtract one method is so nice I discovered it by myself when I was to lazy to do Euclidean division
Me who would solve it breaking x^2+1 into x+i and x-i💀💀
I did exactly that, and got (x+i)ln(x+i)+(x-i)ln(x-i)-2x+C
This is actually a way of connecting arctan with complex logarithms, which are defined in terms of log of magnitude and imaginary part of the angle of the complex number, so it's not necessarily a bad exercise to see what happens then!
@@bfmdsm2020 You can simplify it and get the same result.
Depends on the level of the audience. When I taught techniques of integration in high school none of my students were versed in complex numbers well enough. So IBP would have been the absolute go to for this question. Followed by a 3 sec long division.
That actually works. since Complex numbers have an Euler representation (z = |z| e^(i arg(z))) log(z) can be related to arg(z) and |z|. arg(z) links into arctan etc. On simplification it'll give the same thing
how exciting
Reminds me of Andy Math lol
@@Anmol_Sinhayeah 💀
I think substitution of t=ln(x²+1) works as we will then substitute e^t=y and then do a simple integral.
We can substitute x as tany and use by parts and solve
I got sth in integration methods, when we face sth unsolvable DI method is the best TRY 😊
Aw jeez, I did it using trig substitution caused it seemed easier but then I still had to do integration by parts.
0:50 - will use formula integral I guess?
Just integrate by parts!
I supposed when I saw the question at first!
I did the same thing
I think both of them under-pressure did a calculation mistake... Quite sad, I could see that the first guy was going for integration by parts...
My thought was to integrate ln(x+i) + ln(x-i) dx -> (x+i)ln(x+i) - x - i + (x-i)ln(x-i) - x + i + C
= xln(1+x^2) - 2x + i ln((x+i)/(x-i)) + C
[ i ln((x+i)/(x-i)) = i ln((1-ix)/(-1-ix)) = 2iartanh(-ix) = -2i^2 arctan(x), using identity artanh(ix) = iarctan(x) ]
= xln(1+x^2) - 2x + 2arctan(x) + C
我算對了，我是積分超人🎉🎉🎉
Can you please solve integration 0 to pi/2 ln(1+tanx)/sinxcosx
This integral is divergent.
@@CryToTheAngelzpi not infinity
This function has a vertical asymptote at pi/2
@@ianfowler9340 well you can always ignore a point in definite integration
When one of the bounds of integration is a vertical asymptote it cannot be ignored.
int_0 to 2 (1/x^2) dx
int_0 to pi/2 (tan x) dx
It seems poly long division gets such a bad rap on TH-cam math videos. It takes just a couple of seconds and is so easy and reliable. Adding zero hmmm.