An easier way of getting to the conclusion presented at 4:32 would be to let u = arcsin(x) and let dv = ln(x)/x, which would imply that v = ln^2(x)/2 by using the chain rule in reverse. It also skips the step of computing the derivative of arcsin(x)ln(x) which is a bit more complicated.
@@manstuckinabox3679 Me too, the (1/x dx) with a ln(x) is very suggestive of this approach, not to mention that taking the derivative of arcsin(x) removes its trig nature.
Another faster solution is making subsititution x= sin(t) and the integrand becomes t cot(t) log(sin t) then integration by parts by letting u=t and dv= cot(t) log(sin t) dt, after all we end up in the given integral in first tool.
I think a good topic for a video would probably be what makes a function un-integratable within the set of elementary functions. So for example we know that the integral of e^(-x^2) is non-elementary, but how is this proven? And why this simple looking function can't be integrated, while some crazy ones you have done here can be?
I tried to calculate this integral by series expansion of arcsin(x) (binomial expansion of (1-x^2)^(1/2) ) then i integrated by parts to get rid of ln and have got sum
The *favorite integral suggestor* is some community member who likes suggesting Michael lots of interesting integrals to make videos on. He seems to enjoy this person's suggestions most, hence the name favorite. In fact, there is a google forms document in the description where everyone can suggest Michael problems, although I am not sure if that's actively used right now.
The geometric interpretation of the definite integral is the signed area under the function's graph, where signed area means parts above the x-axis evaluate to positive numbers and parts below the axis evaluate to negative numbers. So in the case of this particular integrant, the negative parts trivially outweigh the positive parts, as the graph is only below the x-axis on the intervall from 0 to 1; actually that appears to be the only place where this function is defined, because of the domain restrictions on ln and arcsin.
∫[0;1] asinx *lnx /x *dx = (*) In parts, and lim(x→0) 1/2 *[ asinx *(lnx )^2]=0. (*) =-1/2 *∫[0;1] ( lnx)^2*dx/ (1-x^2)^1/2 = (**) Replacing the variable t=e^(-t). (**) =(-1/2) *∫[0;∞] t^2 *e^(-t)*[1- e^(-2t)]^(-1/2)*dt =(***) Decompose the denominator into a series of powers of e^(-2*t)
*@ Michael Penn* -- Writing "arcsinx lnx" for the integrand does not make sense. 1) Those supposed functions are separated by a gulf, so they are just "floating" next to each other. There is no operation going on between them. Please use a multiplication dot, an asterisk, or some grouping symbols on your board to indicate an actual product. 2) Those are functions, so the x-variable should have parentheses around it, for example. Please see below for examples of integrands that work: arcsin(x)*ln(x) or arcsin(x)[ln(x)] or arcsin(x)*[ln(x)]
Nice integral
An easier way of getting to the conclusion presented at 4:32 would be to let u = arcsin(x) and let dv = ln(x)/x, which would imply that v = ln^2(x)/2 by using the chain rule in reverse. It also skips the step of computing the derivative of arcsin(x)ln(x) which is a bit more complicated.
lol I was about to drop a by parts the way you suggested.
@@manstuckinabox3679 Me too, the (1/x dx) with a ln(x) is very suggestive of this approach, not to mention that taking the derivative of arcsin(x) removes its trig nature.
Thank you for the video.
6:21 Good day, everyone! Utilize every moment to prepare your mind and body for future endeavors! ✨
Another faster solution is making subsititution x= sin(t) and the integrand becomes t cot(t) log(sin t) then integration by parts by letting u=t and dv= cot(t) log(sin t) dt, after all we end up in the given integral in first tool.
I think a good topic for a video would probably be what makes a function un-integratable within the set of elementary functions.
So for example we know that the integral of e^(-x^2) is non-elementary, but how is this proven? And why this simple looking function can't be integrated, while some crazy ones you have done here can be?
0.41 Don't need Hospital's rule. We knew arcsinx goes to zero more rapidly and faster than (lnx)^2 goes to infinity. So the limit is equal to zero.
Why don't you use that d/dx 1/2ln(x)^2=ln(x)/x to perform the integration by parts?
I tried to calculate this integral by series expansion of arcsin(x) (binomial expansion of (1-x^2)^(1/2) )
then i integrated by parts to get rid of ln and have got sum
Have you not done this integral before, after finding a series for arcsin?
Hello, I'm from Spain, I have a question: can u film a video about int x^2cosh^{-1}x dx
But if you don't know those two previously done integral hints as I assume most ppll don't can you solve without just integration by parts ?
what is "favorate integral suggestor"?
The *favorite integral suggestor* is some community member who likes suggesting Michael lots of interesting integrals to make videos on.
He seems to enjoy this person's suggestions most, hence the name favorite.
In fact, there is a google forms document in the description where everyone can suggest Michael problems, although I am not sure if that's actively used right now.
Sorry what does it mean that the result is negative?
The geometric interpretation of the definite integral is the signed area under the function's graph, where signed area means parts above the x-axis evaluate to positive numbers and parts below the axis evaluate to negative numbers.
So in the case of this particular integrant, the negative parts trivially outweigh the positive parts, as the graph is only below the x-axis on the intervall from 0 to 1; actually that appears to be the only place where this function is defined, because of the domain restrictions on ln and arcsin.
∫[0;1] asinx *lnx /x *dx = (*)
In parts, and lim(x→0) 1/2 *[ asinx *(lnx )^2]=0.
(*) =-1/2 *∫[0;1] ( lnx)^2*dx/ (1-x^2)^1/2 = (**)
Replacing the variable t=e^(-t).
(**) =(-1/2) *∫[0;∞] t^2 *e^(-t)*[1- e^(-2t)]^(-1/2)*dt =(***)
Decompose the denominator into a series of powers of e^(-2*t)
*@ Michael Penn* -- Writing "arcsinx lnx" for the integrand does not make sense. 1) Those supposed functions are separated by a gulf, so they are just "floating" next to each other.
There is no operation going on between them. Please use a multiplication dot, an asterisk, or some grouping symbols on your board to indicate an actual product. 2) Those are
functions, so the x-variable should have parentheses around it, for example. Please see below for examples of integrands that work:
arcsin(x)*ln(x) or arcsin(x)[ln(x)] or arcsin(x)*[ln(x)]