Another method is using integration by parts (DI method), taking x to derive and sin/(1+cos^2) to integrate, so we have: [-xarctan(cos(x))] to evaluate from 0 to pi + integral from 0 to pi of arctan(cos(x)) (the minus before the second integral and the minus before arctan(cosx) are simplified). The first part gives (pi^2)/4, the second integral with substitution u=cos(x) change in the integral from -1 to 1 of arctan(u)/(1-u^2)^(1/2). Since the last one is a definite integral of an odd function evaluated from -a to a, it gives 0. So the final result of pi^2/4
Alternatively just realize that: the integral from 0 to pi of arctan(cos(x))= the integral from 0 to pi of arctan(cos(pi-x))= the integral from 0 to pi of arctan(-cos(x))= the integral from 0 to pi of -arctan(cos(x))= -the integral from 0 to pi of arctan(cos(x)) So the integral is negative itself, so it must be 0.
Seems easier to notice that arctan is odd and cos has odd symmetry about pi/2 and so you get 0 from the "correction". This is a more obvious way to see it than "integrate from left to right and right to left and see the cancellation", I think.
This question appears in my Calculus 2 homework, where a hint is given: Show by substitution that Integral from 0 to pi of x*f(sinx) = pi/2 * Integral from 0 to pi of f(sinx)
Very nice! When it comes to integration techniques, I still have a long way to go. I'll try to add those trig identities and substitution to my toolkit, and I'll keep watching your videos. Thanks for making them!
You know, Michael puts out a lot of videos--which I am very happy with. Honestly, though, I wouldn't mind him live-solving some questions every know and then--especially if he can talk through them too. Actually, it would be really cool if he solved some harder questions that way I (and all his viewers) can experience what it looks like to solve a question. Beats me, but would really love to see that being a thing. Cheers!
I did the u=cos(x) substitution first to give an integral from -1 to +1, which is symmetric about u=0. Then consider whether the the factors of the integrand are odd or even functions. Note that arccos(u)-pi/2 is odd & use this fact to split the integral into a sum of two integrals, both from -1 to +1. The first term is an integral of an odd function times an even, so the integrand is odd. It's integrated from -1 to +1, so it is zero. This leaves the second term, which is a constant (pi/2) times the integral of 1/(1+u^2) from -1 to 1, which is arctan(1)-arctan(-1).
Hi, And if we had x^2 instead of x, should we have (a rationnal multiple of) pi^3 ? For fun: 2:13 : "ok, cool", 3:04 : "1+1=2 , times a half, is clearly 1"
Use the substitution u=cosx. Integral becomes cos^-1(u)/(1+u²) from -1 to 1. Using the identity sin^-1(u)+cos^-1(u)= π/2 for u in [‐1,1], the new integral becomes (π/2-sin^-1(u))/(1+u^2) from -1 to 1. This can be split into 2 integrals, but one of them is 0 due to odd symmetry. The remaining integral is π/(2*(1+u²)) from -1 to 1 which is equivalent to π/(1+u²) from 0 to 1 by even symmetry. That evaluates to π²/4 as desired.
Can you prove this result of Ramanujan involving infinite nested radicals. I wasn't able to find a proof: x+n+a=sqrt{ax+(n+a)^(2)+x(sqrt{a(x+n)+(n+a)^(2)+(x+n)(sqrt{ etc.
Hello Michael, I will start to engineering faculty this year(chemical engineering) I started to Robert Adams Calculus which the syllabus suggests and I am now at page 136. I have a question: I am making lots of proofs each day, I have proven general leibnitz rule for more than 2 functions, derivative of logarithm etc etc. but I wonder if making proofs will get me anywhere in engineering. Will this kind of pursuit in mathematics help any in engineering?
in India for competitive exams , we would have substituted x=t-π/2 in order to obtain the limit -π/2 to π/2 .After that 2 classic identity would have given the solution in 2 lines tops
The function arctan(x) is defined as having a range of (-π/2,π/2), hence why -π/4 was chosen. The bounds of integration are relevant to the numbers being substituted into the function, not the final value of the definite integral itself.
That works too. You run into the integral from 0 to π of arctan(cosx) using that method, which you can evaluate using the substitution u=π-x, and the fact that cos(π-x)=-cosx. This means that the integral is equal to the negative integral, and only one number has that property: zero. There's also a more elegant way to evaluate that integral by breaking up the bounds from 0 to π/2 and from π/2 to π, but I'll leave that up to you. Anyway, you get the same answer, π²/4, after evaluating -xarctan(cosx) from 0 to π.
I don't see why anyine would think pf breaking this up with the one half..it's arbitrary and feels like cheating and probably shouldn't be allowed...don't see why anyone would ever think of it.
Huh, I don't remember the GRE having any integral calculus when I took it. Guess I got lucky, it had been three years since I done any calculus when I took it.
@@rbnn hmmm, I took it in 2015, I think I just got really lucky. Was all computerized, and I don't think I used my scrap paper at all. I'd say they have some balancing issues, if there's people seeing questions like this, mine was all aptitude stuff, very little actual skill and knowledge questions. Just seems kinda unfair to other prospective students, mine was a cake walk.
There is no such rule as King's rule (or Queen's rule if u think that also exists) and also Why do we have to memorise the rules if we can just use our brain to think of some creative approach to the solution rather than filling it with facts and formulas
@@techboyzeeshan8582 yeah why you learn result of gaussian integral When you can proof it That not the right answer King,queen and jack are basic rules
We know that the integeral of sinx/(1+cos^2 x) = arctan(cosx) Using integeration by part D I x sinx/(1+cos^2x) 1 arctan(cos x) dx Our integeral I = x arctan(cosx) - integeral of arctan (cos x) For second part sub u=pi/2- x du=-dx Integeral of arctan (sinu) from u= -pi/2 to pu/2⁹ The finction arctan(sinu) is odd which mean that integeral =0 I = x arctan(cos x) from 0 to pi = pi arctan (cos pi) = pi^2 / 4
Another method is using integration by parts (DI method), taking x to derive and sin/(1+cos^2) to integrate, so we have:
[-xarctan(cos(x))] to evaluate from 0 to pi + integral from 0 to pi of arctan(cos(x)) (the minus before the second integral and the minus before arctan(cosx) are simplified).
The first part gives (pi^2)/4, the second integral with substitution u=cos(x) change in the integral from -1 to 1 of arctan(u)/(1-u^2)^(1/2). Since the last one is a definite integral of an odd function evaluated from -a to a, it gives 0. So the final result of pi^2/4
Nice
Alternatively just realize that:
the integral from 0 to pi of arctan(cos(x))=
the integral from 0 to pi of arctan(cos(pi-x))=
the integral from 0 to pi of arctan(-cos(x))=
the integral from 0 to pi of -arctan(cos(x))=
-the integral from 0 to pi of arctan(cos(x))
So the integral is negative itself, so it must be 0.
Seems easier to notice that arctan is odd and cos has odd symmetry about pi/2 and so you get 0 from the "correction". This is a more obvious way to see it than "integrate from left to right and right to left and see the cancellation", I think.
That’s how I solved it, no time to think
Michael: "Wishful thinking.."
bprp: "Wouldn't it be nice.."
underrated comment
Maybe bprp is just a massive Beach Boys fan?
This question appears in my Calculus 2 homework, where a hint is given:
Show by substitution that
Integral from 0 to pi of x*f(sinx) = pi/2 * Integral from 0 to pi of f(sinx)
right. the quickest way to do it.
@@RL-mv9hj but one has to know this fact a priori to solve in this fashion
I didn't know King's property before, I discovered it with an old video of yours. That is a really trivial fact but very useful, thanks!!
7:58
accurate
not too chatty today?
Congrats to 300k!
Seeing that (pi - x) appear I suddenly realised where he was going, and I just grinned with delight. That was sweet, man. That's what I'm here for.
Very good. That's very neat techniques. Thanks. And explained well, as always.
Integration by parts to get rid of the x and then the remaining integral odd wrt x=pi/2 and is this zero. pi^2/4
This is what I basically did as well
Very nice! When it comes to integration techniques, I still have a long way to go. I'll try to add those trig identities and substitution to my toolkit, and I'll keep watching your videos. Thanks for making them!
@@snehasismaiti342 Wow in what way?
You know, Michael puts out a lot of videos--which I am very happy with. Honestly, though, I wouldn't mind him live-solving some questions every know and then--especially if he can talk through them too. Actually, it would be really cool if he solved some harder questions that way I (and all his viewers) can experience what it looks like to solve a question. Beats me, but would really love to see that being a thing. Cheers!
I did the u=cos(x) substitution first to give an integral from -1 to +1, which is symmetric about u=0. Then consider whether the the factors of the integrand are odd or even functions. Note that arccos(u)-pi/2 is odd & use this fact to split the integral into a sum of two integrals, both from -1 to +1.
The first term is an integral of an odd function times an even, so the integrand is odd. It's integrated from -1 to +1, so it is zero.
This leaves the second term, which is a constant (pi/2) times the integral of 1/(1+u^2) from -1 to 1, which is arctan(1)-arctan(-1).
Hi,
And if we had x^2 instead of x, should we have (a rationnal multiple of) pi^3 ?
For fun:
2:13 : "ok, cool",
3:04 : "1+1=2 , times a half, is clearly 1"
Use the substitution u=cosx. Integral becomes cos^-1(u)/(1+u²) from -1 to 1. Using the identity sin^-1(u)+cos^-1(u)= π/2 for u in [‐1,1], the new integral becomes (π/2-sin^-1(u))/(1+u^2) from -1 to 1. This can be split into 2 integrals, but one of them is 0 due to odd symmetry. The remaining integral is π/(2*(1+u²)) from -1 to 1 which is equivalent to π/(1+u²) from 0 to 1 by even symmetry. That evaluates to π²/4 as desired.
Can you prove this result of Ramanujan involving infinite nested radicals. I wasn't able to find a proof:
x+n+a=sqrt{ax+(n+a)^(2)+x(sqrt{a(x+n)+(n+a)^(2)+(x+n)(sqrt{ etc.
Its easy if you know that integral from x=a to x=b of f(x) is same as integral from x=a to x=b of f(a+b-x)
Strictly that should be: same as integral from x=b to x=a of f(a+b-x).
@@Hiltok No, the order is flipped back to [a, b] again by the negative sign on the dx after substitution
@@flyingpenandpaper6119 You (and Ishtiak) are quite right. Don't know what I was thinking.
Thank you for the video!
Hello Michael, I will start to engineering faculty this year(chemical engineering) I started to Robert Adams Calculus which the syllabus suggests and I am now at page 136. I have a question: I am making lots of proofs each day, I have proven general leibnitz rule for more than 2 functions, derivative of logarithm etc etc. but I wonder if making proofs will get me anywhere in engineering. Will this kind of pursuit in mathematics help any in engineering?
It will help with problem solving for sure, but the actual method of proof writing not so much
Can someone be both an engineer and scientist with double majoe in PhD
@@Abcd-hs4cv it’s certainly possible, but very difficult, especially if you want to graduate in the usual amount of time
in India for competitive exams , we would have substituted x=t-π/2 in order to obtain the limit -π/2 to π/2 .After that 2 classic identity would have given the solution in 2 lines tops
Were the hints included with the problem? I couldn't solve it from just the thumbnail, but once I had the hints, it was trivial. Thanks as always!
Should explain why arctan (-1) is chosen to be -π/4 and not 3π/4. Particularly since -π/4 is outside the bounds of integration and 3π/4 isn't.
The function arctan(x) is defined as having a range of (-π/2,π/2), hence why -π/4 was chosen. The bounds of integration are relevant to the numbers being substituted into the function, not the final value of the definite integral itself.
This is the first problem I solved by myself without looking at your solution :)
Hi,
Why do you use these dummy colors?
Holy it do appears in our past paper (HK student, ref. HKALE P maths 1999/II/Q2b)
Interesing way! But why not use only method by parts? There Is integral arctg(cosx) on interval (0,pí),IT will be zero.
@@DonLicuala Hi, I used method by parts.(x)'=1, then is there(x go till pí) lim(x* -arctg cosx)+ integral(arctg cosx),
Your videos are really helpful🙏🙏.......wish me luck sir i have my JEE EXAM next week....👏👏
How bout doing the indefinite integral?
You can do that using integration by parts (+ the hint), but for the definite integral given, this solution is much cheekier.
It looks like this integral could also have been evaluated using the DI Method given hint 1.
That works too. You run into the integral from 0 to π of arctan(cosx) using that method, which you can evaluate using the substitution u=π-x, and the fact that cos(π-x)=-cosx. This means that the integral is equal to the negative integral, and only one number has that property: zero. There's also a more elegant way to evaluate that integral by breaking up the bounds from 0 to π/2 and from π/2 to π, but I'll leave that up to you. Anyway, you get the same answer, π²/4, after evaluating -xarctan(cosx) from 0 to π.
Current PhD student, would not have been able to do this in the allotted time for an average question for the math GRE 🙃
A reminder:
The video's cover picture has forgotten about the 'dx'
Damn, I had almost forgotten this technique. 😅
Time to revisit the basics, I guess
why so fancy if u can do integration by parts?
lets u = x
dv= sin x / 1+ cos^2 x
basic 💯
I would have spent hours doing IBP, drawing triangles and the like. Never underestimate the power of wishful thinking.
Thank you, professor!
IBP does work here as well, it is somewhat less elegant, but it does work
Just apllying kings rule this becomes pretty simple, u just need to remeber that rule and lots of challenging looking trig integral become easy
There has to be a more transparent way to see the x sinx simply reduces to .5 pi sinx. hmm..
Maths is really beautifull you just need to see it the correct way
I don't see why anyine would think pf breaking this up with the one half..it's arbitrary and feels like cheating and probably shouldn't be allowed...don't see why anyone would ever think of it.
Very interesting prove.
Diviso2
Huh, I don't remember the GRE having any integral calculus when I took it. Guess I got lucky, it had been three years since I done any calculus when I took it.
Wikipedia says the test was changed in 2001 to make it harder. Much harder from the looks of it.
@@rbnn hmmm, I took it in 2015, I think I just got really lucky. Was all computerized, and I don't think I used my scrap paper at all. I'd say they have some balancing issues, if there's people seeing questions like this, mine was all aptitude stuff, very little actual skill and knowledge questions. Just seems kinda unfair to other prospective students, mine was a cake walk.
It's the GRE math subject test for people trying to get into Math grad school. The regular GRE general test is a joke.
@@robertmines5577 ah, that makes more sense. And yeah, I agree.
very interesting
Am I the only one who uses the letter znuts when doing substitutions?
*_Let's try the well know strategy of wishful thinking_*
I wonder why he does not use king rule 🤔
Not well known outside the JEE community ;)
There is no such rule as King's rule (or Queen's rule if u think that also exists) and also Why do we have to memorise the rules if we can just use our brain to think of some creative approach to the solution rather than filling it with facts and formulas
He use king rule but indirectly
@@techboyzeeshan8582 bro it is properties of definte integral and mostly used that why we say that is king rule
@@techboyzeeshan8582 yeah why you learn result of gaussian integral
When you can proof it
That not the right answer
King,queen and jack are basic rules
dx=Nah
In the thumbnail
I use the technique of "wishful thinking" a lot. lol
1+1 = 2. holup...
Ye sab mai chauthi me krta tha.jk 😂
We know that the integeral of sinx/(1+cos^2 x) = arctan(cosx)
Using integeration by part
D I
x sinx/(1+cos^2x)
1 arctan(cos x) dx
Our integeral I = x arctan(cosx) - integeral of arctan (cos x)
For second part sub u=pi/2- x
du=-dx
Integeral of arctan (sinu) from u= -pi/2 to pu/2⁹
The finction arctan(sinu) is odd which mean that integeral =0
I = x arctan(cos x) from 0 to pi
= pi arctan (cos pi)
= pi^2 / 4
Being asian, the question was easy