since michael asked, here's the vid where he calculates the integral of ln(sinx) from 0 to pi/2: kzbin.info/www/bejne/bajMqpWDoKqZmMU :)
@lawliet22632 жыл бұрын
Thanks
@cicik572 жыл бұрын
great, only this stopped me from solving it )
@CM63_France2 жыл бұрын
And that's a good place to pin. Sound recording is better now, I guess, because Michael spoke louder at that time.
@bottlecapbrony3662 жыл бұрын
Finding the old video is left as an exercise for the viewer
@goodplacetostop29732 жыл бұрын
13:03
@barsercan72032 жыл бұрын
I solved it with by making it a double integral first and then changing the order of integration. I chose writing x^2 as integral of 2y from 0 to x and the region in xy plane is a simple triangle and it worked out pretty nicely. I think it's really helpful to consider this strategy if the bounds are finite and nice, because it unlocks the tools available for double and triple integrals. Thanks Michael great video
@mathflipped2 жыл бұрын
Interesting problem! Great job, Michael.
@jesusalej12 жыл бұрын
If you put down one of the x's, you get sinx/x in the denominator whos limit is 1. Above you get x cosx whos limit is 0
@yowaimo8902 жыл бұрын
Ew that's a very generic method let's go with applying first principle ;-;
@giuseppemalaguti4352 жыл бұрын
piln2+(1/12)pi^3
@levanjikiya44542 жыл бұрын
I just multiplied by 1, which is (sin^2(x)+cos^2(x)) and the whole thing was just simple integration from what we learn in Calc 2.
@aadhaarmurty11802 жыл бұрын
You can also proceed by integration by parts if you know your antiderivatives really well. The integral is equivalent to the integral of x^2 csc^2(x) minus pi^3/24. Just differentiate x^2 (x^2 ---> 2x --->2) and integrate csc^2(x) (csc^2(x) ---> - cotx ---> - ln(sinx) ) apply the formula, evaluate the limits and you're done.
@sagarmajumder78062 жыл бұрын
Sir, in second limit just use 1)fundamental theory of engineering is sinx=x, then L(xlnx)=0.(2) when x tends to 0 then sinx= x. For both reasons calculation will be shorter.🙏🙏
@thsand50322 жыл бұрын
A friend gave it to me and I just nuked it with the residue theorem (separate x²/tan²x = x²/sin² x- x², integrate x²/sin²x by parts to get 2x/tan(x), which turns into arctan(x)/x(1+x²) from 0 to infinity. Write arctan(x) = 1/2i [log(1+ix)-log(1-ix)], separate in two, sub in the 2nd to get an integral over R, residue nails it). I just love the residue thm so much
@mryip062 жыл бұрын
4:26 Should we use lim(x->0) x/sin x = 0 ?
@gregsarnecki75812 жыл бұрын
Please tap the blackboard when you make the jump to a cleaned-up half of a blackboard - it enhances your already magician-like status as the maths YT wizard!
@panadrame39282 жыл бұрын
When calculating limits, why don't you use equivalents (dunno if this is the english term) ? Like sin x ~ x and then since it doesn't tends toward 1 when x -> 0, ln(sin(x)) ~ ln (x) then x ln(sin x) ~ x ln (x) -> 0 Quite faster... And smoother It kinda works for any "hard" limit so you should consider using this method
@autumn_auburn2 жыл бұрын
This approach isn't rigorous. You'd probably only use it in exams like SAT where no reasoning has to be provided for the answer and there is a time constraint.
@SuperSilver3162 жыл бұрын
Do u = tanx, then the integral becomes Int(arctan^2(u)/((u^2)*(u^2+1))) from 0 to inf You can use partial fractions to break up the integrals into int(arctan^2(u)/u^2)-int(arctan^2(u)/(u^2+1)) Both of these from 0 to inf Then you can compute these integral individually easily, one by Feynman integration, and the other by u substitution. Pretty cool integral!
@victormd11002 жыл бұрын
How do you propose to evaluate the one by feynmann integration?
@SuperSilver3162 жыл бұрын
Sorry just saw this! You can integrate it parts first and that turns that first integral into int(arctan(x)/((x)(x^2+1))) from 0 to inf the uv parts will vanish due to the limits. Then define I(t) as I(t) = int(arctan(xt)/((x)(x^2+1))) From there take a derivative with respect to t, the x on the bottom gets canceled dI/dt = int(1/((1+x^2)(1+x^2t^2)) From here it’s a partial fractions for the integral, then integrate one more time in the t variable, and you should be able to arrive at the answer.
@victormd11002 жыл бұрын
@@SuperSilver316 thank you, it was really helpful
@holyshit9222 жыл бұрын
I calculated it the same way as Michael but before watching video
@txikitofandango2 жыл бұрын
I can't easily find the video, but with simple u substitutions (u = π/2 - x or u = π - x) you can show that, if I = integral from 0 to π/2 of ln(sin x)dx J = integral from 0 to π/2 of ln(cos x)dx K = integral from π/2 to π of ln(sin x)dx, then I = J = K. Then 2I = I + J = integral from 0 to π/2 of ln(sin x * cos x)dx = integral from 0 to π/2 of (ln(sin(2x)) - ln(2))dx With a simple u-sub u=2x this becomes 2I = (1/2)(I + K) - integral from 0 to π/2 of ln(2)dx I = - integral from 0 to π/2 of ln(2)dx = -πln(2) / 2
@8jhjhjh2 жыл бұрын
Aha that’s pretty neat
@dariusbic2 жыл бұрын
Nice exercise por a Calculus exam!!
@richardheiville9372 жыл бұрын
d/dx tan(x)=1+tan(x)^2 and tan(0)=0 therefore lim x=0, tan(x)/x=1
@tgx35292 жыл бұрын
And what about use the idea(similar) d/dx(g(x)/tgx)=(g'(x) tgx-g(x)(1+(tgx)^2))/(tgx)^2 Then g(x)/tgx=integral g'(x)/tgx dx- integral g(x)/( tgx)^2 dx- integral g(x) dx For g(x)=x^2 we can see there the original integral......
@cernejr2 жыл бұрын
Approx. 0.885658
@GirishManjunathMusic2 жыл бұрын
This video was very fun to watch, and the detective-work challenge was enjoyable to complete, thank you!
@random199110042 жыл бұрын
Just about every time there is an indeterminate form in the 'UV' part of an integration by parts, the answer turns out to be 0. There must be some reason why...
@ConManAU2 жыл бұрын
Mostly because they’re usually a polynomial going to 0 times a more complicated function going to infinity, and those usually come down to the limit being either 0 (if the polynomial wins) or ♾ (if the other function wins), and answers of infinity are often unsatisfying for these sorts of problems.
@belalahmed28262 жыл бұрын
Peace upon you Integral of (tan x)/((cos x sqrt(xsin^5sqrtx)) dx Substitute t= sin sqrtx
@winky321742 жыл бұрын
Nice!
@leif10752 жыл бұрын
Whybuse natural log of sine x at all when there is no log and nonsine by itself here..I don't see why anyone would.think of it..why not natural log of tangent..
@manucitomx2 жыл бұрын
An easy to follow complicated-ish integral. Thank you, professor.
@anunrecognizedgenius33292 жыл бұрын
The hypothesis Andrew Beal, was wrong, 128^5+32^7=8^12, do you want to know more?
@qm_turtle2 жыл бұрын
Your example doesn't violate the Beal conjecture, since 128, 32, and 8 all have a the common prime factor 2.
@anunrecognizedgenius33292 жыл бұрын
@@qm_turtle in general, it violates, another common divisor is the number 4.
@qm_turtle2 жыл бұрын
@@anunrecognizedgenius3329 I don't know enough about the Beal conjecture, so I needed to look it up on wikipedia in the first place. There it is stated for the Beal conjecture that if A^x + B^y = C^z, with x,y,z,A,B,C non-zero integers and x,y,z >= 3, then A,B,C have a common prime factor. There is nothing stated about other common divisors. So how does your statement about the common divisor 4 come into play here?
@anunrecognizedgenius33292 жыл бұрын
@@qm_turtle Wikipedia says that in the hypothesis there is only a simple divisor, as I understand it.
@anunrecognizedgenius33292 жыл бұрын
@@qm_turtle no i sostavnym. I can prove that in the equation, the divisor can be not only prime, but also composite
@沈栋-w9h2 жыл бұрын
Nice!
@MrFtriana2 жыл бұрын
This is the link: kzbin.info/www/bejne/bajMqpWDoKqZmMU
@michelbernard90922 жыл бұрын
TY !
@chmjnationalsuperarmygener85642 жыл бұрын
Hi
@richardheiville9372 жыл бұрын
antiderivative of 1/tan(x)^2 is -1/tan(x)-x and antiderivative of 1/tan(x) is log(sin(x))