Beautiful Integral from Stanford Math Tournament

[@avananana]

Integration is a form of art itself. It's damn beautiful once you get it to work the way you want it to.

[@AayushVarma18]

Someone rightly said differentiation is a technique, but integration is an art!

[@codetrooper9279]

Yes it is really

[@bertiewarner9381]

You can also do the cos equality first to reduce it to the integral from 0 to pi of xsin(x)/(1+(cosx)^2) dx then make the substitution u=cos(x) which gives the integral (after incorporating minus sign into the limits) from -1 to 1 of arccos(u)/(1+u^2) du. We can now go integration by parts recognising that the bottom is the derivative of arctan(u) to get that the answer is (arccos(u)*arctan(u)) evaluated between -1 and 1 plus the integral from -1 to 1 of arctan(u)/sqrt(1-u^2) du. This gives us pi^2/4 plus the integral but we notice that the integrand is odd and integrated between -1 and 1 so evaluates to 0, therefore our answer is (pi^2)/4

[@edmundwoolliams1240]

I did it differently. I used integration by parts on the original integral to get I = (pi^2)/4 + int(0 to pi, arctan(cosx)dx). Then I used u = pi/2 - x to convert it to I = (pi^2)/4 + int(-pi/2 to pi/2, arctan(sinx)dx), then subbed u = -x to get another equation which I could then add to eliminate the nasty int(-pi/2 to pi/2, arctan(sinx)dx) to leave 2I = (pi^2)/2, and I = (pi^2)/4 . But I think your method was much slicker, it was interesting to see how two separate journeys through integration yielded the same correct result.

[@martinepstein9826]

That's what I did, except you can skip that last substitution because int(-pi/2 to pi/2, arctan(sinx)dx) is the integral of an odd function over a symmetric interval and therefore 0.

[@michaelsonntag1469]

int(0 to pi, arctan(cos(x))dx) is zero due to symmetry reasons.

[@pbj4184]

@@martinepstein9826 I mean, that is how the symmetry argument works. He just proved it when he wanted to use it

[@holyshit922]

1. Substitution to get interval symmetric over zero 2. Split integrand to even and odd part 3. Use trig identity to expand cos2x 4. Evaluate integral with even part of integrand with substitution and possibly partial fraction decomposition In fact partial fraction decomposition will not be necessary because we will get arctan

[@minhazulislam4682]

I was stuck as I had an x * sinx. I learn today we can solve this problem by changing variables(which helps simplify). I somehow reached the ending similar to this, just that one x wasn't letting me tan^-1. Good one, thanks for making this one.

[@HisMajesty99]

Wow, beautiful

[@benburdick9834]

I love these integration videos so much. Please, I would love to see more!

[@decentman7555]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@decentman7555]

you can see here a lot of tricky integration problems solving by short cut method.... see this given link.. kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@H2CO3Szifon]

from 7:51: you could also observe that the numerator of the remaining integrand is almost the derivative of the denominator, (give or take a factor of -1/2), so its antiderivative is something like -1/2 ln (3 + cos 2u).

[@ianrobinson8518]

Árpád Goretity No, not quite. Differentiate your “solution” and you will have sin(2u) on the numerator.

[@H2CO3Szifon]

oh, right, it's sin(2u), that might not make things easier, then.

[@ianrobinson8518]

Don’t be embarrassed. I thought I saw the same thing too and was confused for a while how a log could produce the desired result! In fact it produced zero. I eventually tweeked to my error.

[@andrewkreisher689]

i like this one a lot, one day i hope to compete in the mit integral bee so this is good practice

[@ruetata11]

Any student can compete for mit bee????

[@ruetata11]

Please reply...... 🤨🤨

[@Ty-sc2ri]

@@ruetata11 what do you think?

[@ruetata11]

@@Ty-sc2ri IDK, I GUESS ONLY MIT STUDENT CAN APPLY

[@pinsonraphael4873]

Great video! This is a special case of a result for symmetrical functions: if f satisfies f( a + b - x) = f(x) (meaning it has an axis of symmetry at x = (a+b)/2) , then the integral of xf(x) from a to b is equal to (a+b)/2 times integral of f(x) from a to b. Applying this to f(x) = 2sin(x)/(3+cos(2x)) immediately yields the result because it satisfies f(pi - x) = f(x). I think you even proved this result on your channel in another video to solve a difficult-looking integral

[@iamtrash288]

I solved it like this(got bored typing halfway thru so sorry) we could expand 3 + cos(2x) to 2(1 + (cos(x))^2), eliminate the 2 above and below and get the integrand (x*sinx*dx)/(1+(cosx)^2). u = cosx => du = sinx*dx (we have it already) u(pi) = -1 u(0) = 1 so we have a definite integral from 1 to -1 of arccos(u)/(1+u^2) integrating it by parts we get - integral from -1 to 1 of arctan(u)/sqrt(1 - u^2) but notice that atctan is an odd function while the expression below is even. Add the fact that we have a symmetric interval and we get that this integral is actually equal to zero and we get the answer (written in blabla)

[@HeyKevinYT]

okk

[@AakashKumar-by6gi]

Your maths is beautiful. I'm in love with it!!!!!!!!!!!!!

[@rickybobby5584]

nice and by the way because the integrand is an even function, we can find the integral for limits from 0 to pi/2 by dividing the final answer by 2

[@terry_tamali]

You can simply apply the limit formula of definite integrals, get rid of the x, convert cos2x into 2 cos^2x -1, take 2 outside from the denominator, then substitute t for cosx - there ya have it!!

[@closed1209]

omg this is exactly one of my homework , you save my life thank you so much

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐💐

[@duncanw9901]

I often wonder on these tests if they would allow you to solve them using laplace-inverse-laplace stuff.

[@decentman7555]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@maalikserebryakov]

What do you mean

[@outofbox000]

Man that was a beauty.

[@sunitajha7596]

Well, it was easy. But very elegant. I initially tried integration by parts, but the problem became trivial when I replaced x by π - x and added.

[@dydx3741]

really amazing ... so soothing and satisfying

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@theoleblanc9761]

I did it like that: first remplace cos2x by 2cos^2x-1, you get: int btw 0/π of xsinx/(1+(cosx)^2) Sub cosx by u you get: int btw -1/1 of arccos(u)/(1+u^2) Integration by part and you get: arccos(u)arctan(u) btw -1/1 + int btw -1/1 of arctan(u)/√(1-u^2) the intergral part is 0 because it is odd, Finally you get π^2/4 👍 Really nice integral! Thank you! Your method is really nice, it is really important to see symmetries like this one on cosine and sine functions!

[@wedemeyermath4896]

When I first came across this problem, I uploaded a solution video as well. This is one of my favorite integrals :)

[@dydx3741]

yes i watched your video :)

[@computergenius365]

This is so beautiful

[@shenal8167]

This is beauti of maths

[@silentintegrals9104]

totally agree!

[@tgx3529]

Integral= integral x* sinx/(1+ ( cosx)^2) dx on (0;π). Primitive function to sinx/(1+(cosx)^2) is function - arctg( cosx). Then se can use metod by parts [ x*(- arctg( cosx)] + integral arctg( cosx))dx , this integral Is eq. zero on(0;π). We get result π*π/4.

[@نعمللوحدة]

Beautiful

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐💐

[@anshumansahu1087]

the day you learn to apply integration to real life problems you can say u have understood integration properly. Till then keep practicing as you still have a long way to go.

[@LetsSolveMathProblems]

Although mathematics is certainly valuable in many areas of sciences (notably physics), I would like to respectfully disagree with your assertion. Mathematics should not be treated as a foundation only existent to merely allow other scientific fields to flourish; rather, it should be seen as a picturesque garden harboring and giving birth to its own flowers of elegance. Perhaps the flowers blossoming in beauty are not necessarily applicable to life--yet, who would dare to argue they should be ignored and thrashed when they are inherently a form of art? Take set theory for example. So far, physics has yet to illustrate the direct physical application of set theory. Nevertheless, set theory forms the ultimate foundation for modern pure math! Not only that, set theory has given birth to a myriad of elegant suppositions about infinity. Same applies for integration. Sure, the nature of integration (that it reverses the rate of change process) makes it an important tool for modeling real life phenomena (Maxwell's equations, displacement, impulse, work, divergence theorem, and many, many more physics laws and equations are based on integration). Yet, just purely by itself, integration is beautiful, exciting, and fun. In the end, the joy of pure math comes from, of course, the joy of it by itself, not necessarily from its connection to real life. Once one truly appreciates the intricate nature of integrals (and other mathematical manifestation of universe and human thought), I think one has accomplished a long journey, not having "a long way to go" in the sense of not having "understood it properly."

[@anshumansahu1087]

LetsSolveMathProblems well I think you have taken this statement of mine as a dig on you but it is not so. It is for all of us in general. Sorry if I have caused any offence.

[@LetsSolveMathProblems]

I apologize if I sounded too sensitive or critical; that was not the intention. I personally have always pondered the relationship between applied and pure mathematics, and I thought you provided a good stimulus for a friendly discussion. I appreciate commenters like you who go beyond the topic at hand to discuss a more general picture. And, as you astutely remarked, it is true that for many viewers (especially those with certain occupations) that applied side of integration is more valuable. Again, I apologize if my tone came out to be censorious. Thank you so much for sharing your opinions! =)

[@YASHGUPTA-ft7mz]

You can use the fact that integral.f(x) ( a to b )= integral.f(a+b-x) this would surely cut x and we will end up a simple trignometric integral . :)

[@MiguelFernandez-ix8mi]

Awesome video! Thank you!

[@decentman7555]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@pooydragon5398]

It's pretty easy if you know that integral of f(x)dx from a to b = integral of f(a+b-x)dx from a to b. Then just do what he did. I learnt it as first property of definite integration.

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@0-1min11]

IIT students can solve it without touching the pen . Experimentally determined.

[@vanchhitdubey]

Use property I(x) = I(π-x) U will get one more I on the RHS with a simple function to integrate Write cos2x = 2cos^2x -1 Take 2 common and cancel it Take cosx =t substitution Integrate easily integration will be tan^-1x , limit changed to 1 to -1 Put limit and finally I = pi^2/4

[@sauravthegreat8533]

Couldn’t you use the formula where if the numerator is a differential of the denominator, the integral of that function is ln(denominator), after making some changes of course like multiplying by -2?

[@tukarampawar4302]

Apply kings property and then substitute sine as t

[@anilsharma-ev2my]

What is relationship between function and it's intergreled equation So we find a simple percentage and get rid of lengthy calculation

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐💐

[@AliAbdalla]

(pi^2)/8 numerically

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐💐

[@willli9529]

i solved it using kings property, first my trigonometrically simplifying the integral to I = int from 0 to pi of 2xsinx/(3 + 2(cosx)^2-1) dx or int from 0 to pi of xsinx/((cosx)^2+1) dx. applying kings property this becomes I = int from 0 to pi of (pi-x) sin(pi-x)/((cos(pi-x))^2+1) dx, or simplified trigonometrically to I = int from 0 to pi of (pi-x) sinx/((cosx)^2+1) dx. adding the original I to the version with king's property applied yields 2I = int from 0 to pi of (x + pi -x)sinx/((cosx)^2+1) dx) which equals int from 0 to pi of pisinx/((cosx)^2+1) dx. from here it is just a u-sub with u = cosx and du = -sinx dx, and the evaluated indefinite integral is pi* -arctan(cosx). plugging in bounds of integration yields 2I = -pi * (-pi/4-pi/4) or (pi^2)/2; thus I = (pi^2)/4

[@Fernando31611]

My Answer: I= Int(0,pi,(xsinx/(3+cos^2(x))= Int(0,pi,(x*sinx/(3+cos^2(x)))= Int(0,pi, x(-arctan(cosx))-int(-arctan(cosx))). Now, I was wondering if (-arctan(cosx)) was simetrical within the range... cos(0) = 1, cos (pi/2)=0, cos(pi)=-1, and is SIMMETRICAL to 0 (with negative values, arctan(-x)=-arctan(x)) then the function is simetrical to pi/2, then int(0,pi,-arctan(cosx))=int(0,pi/2,-arctan(cosx))-int(pi/20,pi,-arctan(cosx))=0. This simplifcation leaves I=(0,pi,x(-arctan(cosx)) = pi*(-arctan(cos(pi))=-pi*arctan(-1)=pi^2/4 Is the symetry assumption correct or just a happy coincidence??

[@ChaineYTXF]

Neat. You just got a new subscriber. :)

[@edwardlewandowski5473]

Faktycznie piękna 📯Bóg jest doskonały w Swoich prezentach🙏🕊️📯

[@Mathelite-ii4hd]

great explanation...loved it...

[@decentman7555]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@donielf1074]

I must be missing something. The value of the integral is the same even though x=π-u? Why can you just set the integrals equal to each other?

[@AKSatMusic]

Because it's definite. You don't have to change the u back to x in your answer because you'll get a numeric answer. If the limits are the same u put the same values in for the variable, giving you the same answer regardless of the variable

[@anilsharma-ev2my]

Which figure fulfilled this integral Like shapes of circles and other geometry possesses different curve but original equation related to them draw themselves Which equation gives right angle triangle after integrating 😁😁😁😁😁

[@mohdahtishamkhan360]

this is an easy problem for those who knew integral property.the property says that, Integral of a function F(x) from X=a to X=b is equal to integral of F(a+b-x) from X=a to X=b

[@BloobleBonker]

Ingenious!

[@anweshaguha7366]

It's the same as King's property isn't it?

[@decentman7555]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@sonamthisside]

I have watched almost your all vedios. I don't understand why do you not use the properties of definite integration in such integration in your vedios

[@decentman7555]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@duncanw9901]

Pro tip: trig functions follow logarithmic arithmetic Actually did that on an Calc 2 test once...

[@JalebJay]

Are you referring to the strategy of subbing all forms of trig to be the form of e^(ix)?

[@Ruby-eq1qg]

Can you elaborate on that a bit?

[@JalebJay]

While I couldn't find a way to use it on this problem. Questions using trig can be solved by replacing cos(x) with Re(e^(ix)). Simple example would be: int(e^x*cos(x)) =Re(int(e^x*(cos(x)+isin(x))) =Re(int(e^x*e^(ix))) =Re(int(e^(x(i+1)))) =Re(e^(x(i+1))/(i+1))+C =Re(e^(x)*(cos(x)+isin(x))*(i-1)/2)+C =Re(e^(x)*(cos(x)+sin(x))+i(sin(x)-cos(x))/2)+C =e^(x)(cos(x)+sin(x))/2+C

[@duncanw9901]

joking a bit... got confused once during an exam

[@vonneumann3592]

Love your vids

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@johnhwhittaker6005]

Holy shit! I was able to solve it on my own for the first time 😭😭😭

[@janmeetsingh7515]

I solved this in less than 2 minutes using properties of definite integration!

[@mike4ty4]

Neat little problem. I solved it in about 1.5 ks or so just before watching the video and you did it mostly the same way, except that I knew immediately to go for cos^2 because I saw a number plus cos(2x) and thought to save a term of 1 from the 3, to get 2 + (1 + cos(2x)) and then that ever-lovin' cosine-squared identity is very conspicuous, you get 2s all over the place, they drop out, and then it's just a matter of noting the bounds and noting that once you put the x aside you are left with a trig bit that under those given bounds has a nice symmetry.

[@markusladen1391]

In Russia we call it homework exercise

[@vineetsai9971]

this is such a easy problem........a TRUE JEE aspirant will surely know this..... :)

[@moussalasfar1512]

Good

[@sagartirthasengupta852]

I gave the 1000th like

[@You12783]

I remember solving it in my board examination ...used the same steps 😁

[@sjsjjf8feirbfjtjfjifofofof417]

Don't lie

[@ГеройАлександрНевский]

@@sjsjjf8feirbfjtjfjifofofof417 hahahaha

[@sjsjjf8feirbfjtjfjifofofof417]

@@ГеройАлександрНевский oh i was hoping it was the original guy providing a sarcastic response :(

[@AnkitKumar-im4yt]

It was an easy one but you are doing an amazing work hope your channel gets it's due recogniztion

[@linguafranca7834]

Beautiful integral 😂😂👌👌☺️

[@shikhanshu]

so freaking deep

[@AimonsL_oignon]

Commenting without seeing is the answer π^2/4? OMG It's right!! My board exam(India) just finished and I wasn't even so delighted then!

[@CheapPhysics]

I have made this man!!!!!!! Give me a heart ...... Please

[@gaurivarshney7971]

Got it right

[@lp4162]

I solved it in 20 sec just by merely looking at it

[@kanishk2345]

Evaluate: ln e

[@TechToppers]

Pi²/4... I'm smelling π²/6

[@rohanroy9329]

This one's pretty simple

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@mohdtauheedkhan6198]

Most easiest problem by using properties of integration directly

[@xaxion_faza2453]

"Sin of you"

[@TRIDEXisFun]

King rule⚠️

[@pyrotas]

Unnecessarily complicated, although informative. Integration by parts is usually the way to go with forms like x^n * {cos,sin}(x). It also leads to recurrence equations for I and yields a faster solution in my opinion.

[@decentman7555]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@dssingh3147]

I am from India and I solve this problem in just one min.

[@silentintegrals9104]

nice!! 😎

[@anilsharma-ev2my]

Graphically Euler equation show what ?😁😭😰

[@sachitananda4684]

easy for a class 12th cbse student

[@nachiketsharma4507]

I solved this in less than 3 minutes

[@rudeviper]

It was a 10 sec problem

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@vonneumann3592]

I will be very much grateful to you if you increase the frequency of your video

[@mohdtauheedkhan6198]

In Allen's sheet

[@radhavenkatramanan7001]

Bro it's too easy. Let's have a harder one next time.

[@decentman7555]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@xSimonTan]

what

[@gardening_vibes]

NCERT Example class 12, interesting

[@pranav7471]

This is too basic, Beautiful WTF?? Check out MIT integration BEE if u want to see beautiful integrals.

[@harshbaliyan5867]

yup property 4 of definite integral is used here

[@domc3743]

you used kings property without explicitly mentionting it, sad face

[@arpit9134]

Not very tough for a serious Jee Aspirant

[@pranab9465]

Dude i solved it on paper under 1min

[@claytoncoe838]

[X] Doubt

[@outofbox000]

I can sense you are an Asian.

[@nicholasheilig3694]

His voice is so annoying

[@edwardlewandowski5473]

Faktycznie piękna 📯Bóg jest doskonały w Swoich prezentach🙏🕊️📯