you can solve directly by changing variables , like it is 0 to pi/2 then you can change X into (0+pi/2 )- X ,we can do this for all function and all integrations so it will be the same integral with different look as you got it will be in form of tan(pi/2-X)^root2.you will directly jump to the point in 2nd step.it will be easy and time saving
You can simply express tan(x) as sin(x)/cos(x) where note that sqrt (2) is a distractor then express our integral as I,then using our reflection formula (integral of f(x) from a to b is equal to the integral of f(a+b-x) from a to b),summing them and divide by two gives the answer!
It can be solved more generally and easily if you write tan=sin/cos and the consider x=pi/2-y to obtain 2I=\int_0^pi/2 1 and then I=pi/4. This result is independent of the power over the tan in the denominator
It can be solved easily using property of integrals as x => upper limit +lower limit -x and then summing the initial integral with the new one as 2I and then it will be integration of dx/2 from 0 to pi/2 I think this is a working approach as well but I do like your approach it is more Putnam style
Being a Calc 2 student who is just fed integral properties, this was really helpful in making me realize that the sqrt is just a distraction and the same thing can be done to any exponent!
Yooo I saw this same integral a couple of days ago and I memorised the thumbnail and tried it. I failed miserably but I couldn't find the fucking video with the solution again. Glad to see it called tour your attention too and that you have it solved for us
*Don't ignorantly and uncivilly curse* here because you were frustrated. It is inappropriate. You could go back and edit it again to eliminate and remove your unnecessary trash.
super hard to do without thinking of the trig identity(which i didn't remember lol) but i guess that's why it's in putnam also for some reason integral calculators don't give an answer-presumably because ''normal'' methods can't be applied to solve it
@@anshumanagrawal346 well yeah probably but then you can only solve it by finding the anti derivative or by approximation through the graph and that's basically guesswork
You won't believe I just solved this problem just hours before the video. I am class xii student from India so this is a easy one for me. Love your work!
Nothing freaky about tricks... If they gonna don't give integrals without Tricks, students have to spend hours to crack them and we can find new Scientists like Einstein or Newton or may be from one of us👍👍👍
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huh, the square root 2 was a complete distraction. great explanation!
could be solved in a more general way for an exponent ''u'' any of tangent, always the result pi/4
Kind of the same, but more Calc 2 oriented :
Use the (b+a)-x substitution trick, then add the integral with it's new version
The sqrt(2), although a distraction, might actually be a hint. If you put a more reasonable exponent, you are tempted to try something else.
One special thing about this integral is that the same trick is also helpful for arbitrary powers of tan x. A very nice problem indeed.
Yes, ur method was less straight forward if one didnt knew kings rule but very elegant!
you can solve directly by changing variables , like it is 0 to pi/2 then you can change X into (0+pi/2 )- X ,we can do this for all function and all integrations so it will be the same integral with different look as you got it will be in form of tan(pi/2-X)^root2.you will directly jump to the point in 2nd step.it will be easy and time saving
You can simply express tan(x) as sin(x)/cos(x) where note that sqrt (2) is a distractor then express our integral as I,then using our reflection formula (integral of f(x) from a to b is equal to the integral of f(a+b-x) from a to b),summing them and divide by two gives the answer!
I solved this using the same method.
Can you elaborate on that
You dont have to Substitute in the first place, just aplly kings rule, add them up and simplify to 2I=pi/2
@@fix5072 I was thinking of the same thing, can you please explain your approach
Blackpenredpen🙌🙌
It can be solved more generally and easily if you write tan=sin/cos and the consider x=pi/2-y to obtain 2I=\int_0^pi/2 1 and then I=pi/4. This result is independent of the power over the tan in the denominator
It's easier when u use definite integral properties
Since tan(pi/2-x) = cot(x)
Hence 2I= pi/2
Hence I=pi/4
I quite liked the approach.
Thank you, professor!
Would love if you could solve recent Putnams
This is a nice problem. Thank you, Michael.
It can be solved easily using property of integrals as x => upper limit +lower limit -x and then summing the initial integral with the new one as 2I and then it will be integration of dx/2 from 0 to pi/2 I think this is a working approach as well but I do like your approach it is more Putnam style
"Some sort of trick" Exactly
Putnam Integrals are really amazing
Being a Calc 2 student who is just fed integral properties, this was really helpful in making me realize that the sqrt is just a distraction and the same thing can be done to any exponent!
Neat integral with a simple and satisfying solution!
KINGS rule also works easily
That last simplification wow
Just amazing !
Yooo I saw this same integral a couple of days ago and I memorised the thumbnail and tried it. I failed miserably but I couldn't find the fucking video with the solution again. Glad to see it called tour your attention too and that you have it solved for us
*Don't ignorantly and uncivilly curse* here because you were frustrated. It is inappropriate. You could go back and edit it again to eliminate and remove your unnecessary trash.
What the fuck bruce i dont know what you talking about. please consider not everybpdy is a native speaker
@@forcelifeforce I'm not frustrated I was grateful to see the solution. grateful and excited!! :)
@@forcelifeforce i just wanted to enfsize that it was a bummer that i lost the original video with that info
@@forcelifeforce oh shit, i understand know why you said frustrated, that kind of frustrated
super hard to do without thinking of the trig identity(which i didn't remember lol) but i guess that's why it's in putnam
also for some reason integral calculators don't give an answer-presumably because ''normal'' methods can't be applied to solve it
Try solving it while proving/justifying every step, you'll realise it's not doable
@@anshumanagrawal346 what do you mean? He justified every step
@@paokaraforlifeI mean without using tricks like substitution, change of variables. Try to solve it as you're trying to find the area under curve
@@anshumanagrawal346 well yeah probably but then you can only solve it by finding the anti derivative or by approximation through the graph and that's basically guesswork
That’s the same result as A6 on the latest Putnam…
that's weird it does not depend on the sqrt(2)
Definite integral questions are never tough
Most of them time use some property and boom you Nr Dr gets cancelled
6:28 A good place to stop.
wow !! The chalk is loud in this video !
That was slick
it was just a week ago or so that this integral appeared on another channel, so...
I feel like there's nothing special with the sqrt(2) power. Any real power a would work
yup
Great effort, using definite properties, would solve it in less than step and within 15 seconds at the most😉😇
Try RMO questions from any year after 2014
Or better- Inmo questions
POV, your looking for an Indian dude explaining how its a 2 second solve using the "kings rule".
that was really cool
So the sqr(2) is used nowhere here... result is the same whatever you put in the exponent of tan(x).
Just apply kings property
Kings Rule!
HELP ME please int ln(sin(t) from x=0 to /sqrt(pi/2)
I think it’s not a trick , just a substitution.. good job btw..
This Trick Was Mind Blowing! I Am Grade 8 But Still Understand This Solution ('Cause Of You). Thank You Sir.
You won't believe I just solved this problem just hours before the video. I am class xii student from India so this is a easy one for me. Love your work!
Same brother
Just kings property!
I know you can....Emmm would just prefer 2020 Putnam competition
Can you end a video with "that's a bad place to stop" once?
When you realize that it came from the Letsthikcritically video
Yep
First
Nothing freaky about tricks...
If they gonna don't give integrals without Tricks, students have to spend hours to crack them and we can find new Scientists like Einstein or Newton or may be from one of us👍👍👍
even you change √2 into an arbitrary number it still works
so we can use e or π too? wow!