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first

Please Integrate x^2 root under ax^2+b by Algebraic manipulation.

using a linear U sub and direct application of the beta function has been done before on this channel, i have some advice that i think i said, if it does not have a unique solution development different from your previous videos, you are basically just posting the same thing, even if numbers are modified

Beautiful!

th-cam.com/video/Ry5q4NsZDx0/w-d-xo.html

Fantastic

sick

Although the solution given in the video is by far the simplest, it may be of interest for some to see how Feynman's trick can apply to this question. Rewrite the original integral as integral_0^inf{dx (ln(1+x^3)-ln(1+x))/(1+x^2)/ln(x)} = F(3)-F(1), where F(a) = integral_0^inf{dx ln(1+x^a)/(1+x^2)/ln(x))}. Therefore, F'(a) = integral_0^inf{dx x^a/(1+x^2)/(1+x^a)} = integral_0^(pi/2){ds tan^a(s)/(1+tan^a(s))} = integral_0^(pi/2){ds sin^a(s)/(sin^a(s)+cos^a(s))} = integral_0^(pi/2){ds cos^a(s)/(sin^a(s)+cos^a(s))} = 1/2*integral_0^(pi/2){ds (sin^a(s)+cos^a(s))/(sin^a(s)+cos^a(s))} = pi/4. Thus, I = F(3)-F(1) = integral_1^3{da pi/4} = pi/2

why make a series of sine and not the e^-x thing

Not sick i got this

Hello, could you please take a look at this integral I've been struggling with: integral from 0 to infinity of sin^2(x)/[x(1+x^2)] dx? Any help would be appreciated.

Good idea. I solved it the same way

Fantastic! What's the title of this magnificent song? You told me TH-cam posts this information a few days after the video upload, but this wait would take too long as the song is too good.

Fantastic

Excellent

This is insane.

Cool

Alternatively, use Feynman's trick. Define F(a) = integral_0^inf{dx sin^2(x) e^(-ax)/x}. Then F'(a)= - integral_0^inf{dx sin^2(x) e^(-ax)} = 1/2*integral_0^inf{dx (cos(2x)-1) e^(-ax)} = 1/2*Real{integral_0^inf{dx (e^(2xi)-1) e^(-ax)}} = 1/2*Real{1/(a-2i) -1/a} = 1/2*(a/(a^2+4) - 1/a). Since F(inf)=0 and I=F(1), we fave I = - integral_1^inf(da F'(a)} = -1/2*integral_1^inf{da (a/(a^2+4) - 1/a))} = ln(5)/4

Cool

Sweet!

this is THE easiest ever on your channel.

Cool

Fantastic

Cool but extremely hard to see

I can offer a more comprehensive solution. Suppose f = cosx+x sinx and g = sinx - x cosx then f`g-g`f = x^2. It means that I = (f`g-g`f) /f^2 => int d(g/f) = g/f + C = (sinx-xcosx) /(cosx+x sinx) +C 👍

insane summation

There's a mistake it must be 1/4√3 tan^-1(2t-1/√3) instead of 1/2√3 tan^-1(2t-1/√3)

It is a very hard problem, and your solution makes it look easy. Hindsight 20-20.

Nice

Similar derivation can be made without invoking the zeta function. First, expand the denominator into a geometrical series. Then the original integral becomes sum_(n=1)^inf{1/(n^2+1)}. This commonly seen sum is usually further evaluated using the expansion pi*cot(pi*x) = 1/x + sum_(n=1)^inf{2x/(x^2-n^2)} and letting x take the value of i. The final result follows naturally.

you can turn the 1/(1-e^-x) to an infinite series and arrive at the same result by using the laplace transformation of sin(x) and the derivative of the infinite sine product

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not sick u already made similar videos

not sick i got this

Perfect

can you do exp(-x)sinx/(1-exp(-x))

I decided the same way 💪

ez saw that coming cuz cosine is above 0 from 0 to pi/2 and then its below until pi and it will always cancel out

i made you derive a formula for this if you remember

similar to what i had told you to calculate a while ago

Nice

try: X double dot = a X + b Y dot; Y double dot = c X dot +e Y. (Second order differential equations with coupled first order derivative terms). Usually solve by reducing to 4 equations and finding eigenvalues, etc. Laplace transform, I don't know?

0 before even watching vid

Nice!

Nice

nice

yes uneasy music fits well a mental integral what is this song

Excellent

do int 0 to inf lnx/sqrtx(sqrtx+1)(sqrt 2x+1)

hm i saw similar integrals lol