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
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.
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
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.🙏🙏
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!
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
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
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.
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!
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.
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......
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..
@@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?
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...
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.
since michael asked, here's the vid where he calculates the integral of ln(sinx) from 0 to pi/2: th-cam.com/video/7wiybMkEfbc/w-d-xo.html
:)
Thanks
great, only this stopped me from solving it )
And that's a good place to pin. Sound recording is better now, I guess, because Michael spoke louder at that time.
Finding the old video is left as an exercise for the viewer
13:03
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
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.
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
Ew that's a very generic method let's go with applying first principle ;-;
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.
Interesting problem! Great job, Michael.
piln2+(1/12)pi^3
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
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.🙏🙏
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!
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
Aha that’s pretty neat
4:26 Should we use lim(x->0) x/sin x = 0 ?
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
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.
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!
How do you propose to evaluate the one by feynmann integration?
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.
@@SuperSilver316 thank you, it was really helpful
Nice exercise por a Calculus exam!!
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......
I calculated it the same way as Michael but before watching video
d/dx tan(x)=1+tan(x)^2 and tan(0)=0 therefore lim x=0, tan(x)/x=1
This video was very fun to watch, and the detective-work challenge was enjoyable to complete, thank you!
Peace upon you
Integral of (tan x)/((cos x sqrt(xsin^5sqrtx)) dx
Substitute t= sin sqrtx
An easy to follow complicated-ish integral.
Thank you, professor.
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..
The hypothesis Andrew Beal, was wrong,
128^5+32^7=8^12,
do you want to know more?
Your example doesn't violate the Beal conjecture, since 128, 32, and 8 all have a the common prime factor 2.
@@qm_turtle in general, it violates, another common divisor is the number 4.
@@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?
@@qm_turtle Wikipedia says that in the hypothesis there is only a simple divisor, as I understand it.
@@qm_turtle no i sostavnym.
I can prove that in the equation, the divisor can be not only prime, but also composite
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...
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.
Nice!
Approx. 0.885658
Nice！
This is the link: th-cam.com/video/7wiybMkEfbc/w-d-xo.html
TY !
antiderivative of 1/tan(x)^2 is -1/tan(x)-x and antiderivative of 1/tan(x) is log(sin(x))
Hi
I am definitely not first
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