Fun side note : There's a nice and quick geometric way to see what A and B are in the equation (sin x + cos x) = A cos (x - B). Consider a square of side 1 unit and a line outside the square. The projection of the adjacent sides of the square on the line is (sin x + cos x) , which is also equal to the projection of the diagonal of the square on the line. The maximum value of the projection is clearly equal to the length of the diagonal or sqrt2 and this occurs when the angle x is equal to pi/4( since the diagonal is parallel to the line in that case ). Therefore A=sqrt2, B = pi/4
From how he said "a Calc 2 student would try..." I kept thinking the trig substitution was going to hit a wall and he'd have to use Feynman's technique
Shorter method : We know integral fx from a to b is same as integral f(a+b-x) from a to b once we got I=ln(1+Tanx) I=ln(1+(tan(45-x)) .. expand tan(a-b)=(tana-tanb)/1+tanAtanB =Ln(1+(1-tanx)/1+Tanx) =Ln(2/1+Tanx) I=ln(2)-I I=1/2ln(2) Now substitute limits I= (1/2)ln(2)*(π/4-0) Done
I didn't really find a way to solve this integral without using advance techniques, and you've just showed you're solution like a champ. I'm shookt. Here's a quick sub for ya!
Literally man, you made this question super tough. Just apply ∫f(x)=∫f(a+b-x) for limits a to b. Apply formula of tan(A-B) for A=π/4 and B=0. Add both equations and apply ln(a)+ln(b)=ln(ab). Only ln2 will left. Answer would come (π/8)ln2. Seriously this is a question from one of the easiest books of Indian Mathematics, NCERT.
wonderful explanation bruh! I have seen lot of your math videos ,It actually helped a lot. :) I am in my second semester currently , you are far better than my math prof. THANKS A LOT KEEP IT UP
i’m loving your videos! super well put together! great content, great editing, and you explain things really well, you’ve got yourself a new subscriber
A very elegant and analytic approach! However, I prefer Feynman's trick in this case. Anyway: I love your condensed and concise way to present the solution.
Thanks for making and sharing another phenomenal video! I was curious: did you know that you needed to end up with the two equal and opposite integrals near the end of the problem, or was that something you stumbled onto or found in a solution writeup somewhere? Either way I think that's amazing; I was just curious about the thought process behind it and how this could help an overall approach to integrals like this. Thanks!
Another method: when we are at ln(1+tan(a)) [let a = theta], we can use king's rule to get ln(1+tan(pi/4 - a)). Next, we can use the tan(a-b) identity, and it will turn out to be ln(2/1+tan(a)). Next, we use the quotient property of ln to separate the integrals and turns out we get the original integral. therefore, assuming the original integral to be I, we get 2I = pi/4*ln2, which evaluates to pi/8*ln2.
there is a much simpler way to solve it. When you have the ln(tan(theta)+1)) just use king property to get ln(tan(pi/4-theta)+1) and it just simplify like butter. You can also try x = (1-t)/(1+t) it work good with log.
Thanks for explaining the sin x + cos x identity! This is my new favorite formula now... why didn't they teach it in schools? (no I am not a JEE aspirant)
It will be much more easier Put I =lim 0 to π/4 ln (1+tan theta) = lim 0 to π/4 ln{1 + tan (π/4- theta) } Next put the formula of tan (π/4-theta) Then I = ln (2/(1+ tan theta)) 2I = lim 0 to π/4 (ln 2) And thus I = ln 2 by π/8
This problem was a regular problem in our textbook. Another easier way would be just to use proven property of a definite integral. But got a new way to solve. Thanks! This was interesting ☺️
Oh why couldn’t this have realised earlier when I had to do this question for my Australian hsc maths ext2 assignment. Just kidding great video, alternate format which I used however was to use the law that you used at the end of the video when you have the integral of ln(tan(theta)-Pi/4), where you can use the tan compound angle formula and do some simplifying in order to get 2x your original integral as equal to pixln(2)/4, giving the same answer as you showed here
After substituting x=tan(theta), Substitute theta=pi/4-T. To rewrite the integral then use the tan(A-B) identity then use the log properties to see the magic happen you would right away get your answere
Well, it literally looks really similar to one of the questions in HKDSE 2020 Maths Exam (The HKDSE examination is the university entrance examination for all high school students in Hong Kong)
I ran into a question like this yesterday, except it also involved partial fraction decomp, and when the terms started canceling out like that I though I did something wrong. And went about it a little different. Now I’m gunna have to go back and look to see if I would have gotten the same answer.
Some time ago i saw integral Int(x/sqrt(e^{x}+(x+2)^2),x) Try to calculate it, and it is not nonelementary as you may think (I can do it , my way involves tricks like adding zero , multiplying by one and then doing obvious substitution) CAS programs like Mathematica or Maple are unable to calculate integral i gave I prefer u=(1-x)/(1+x) substitution for your integral
There also you can do another method by not opening tan into sin and cos and only solve in tan by identity tan a + tanb. Then seperating them and put the limits
The solution for x^2 = 2 is +- root 2 and not just the positive. If the answer was carried out you'd end up with two answers, being the positive and negative version of the one you just got.
@@BriTheMathGuy just looked for this in Wolfram Alpha and it doesn't match www.wolframalpha.com/input/?i=int+from+0+to+1+of+ln%28x%2B1%29%2F%28x%5E2%2B1%29
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Equation solving like this makes me love maths so so much
me too!!
Ah this reminds me of my innocent days when I though all of math was this elegant.
Fun side note :
There's a nice and quick geometric way to see what A and B are in the equation (sin x + cos x) = A cos (x - B). Consider a square of side 1 unit and a line outside the square. The projection of the adjacent sides of the square on the line is (sin x + cos x) , which is also equal to the projection of the diagonal of the square on the line. The maximum value of the projection is clearly equal to the length of the diagonal or sqrt2 and this occurs when the angle x is equal to pi/4( since the diagonal is parallel to the line in that case ). Therefore A=sqrt2, B = pi/4
Love this method, way easier than using the Feynman technique to do it x3
Glad you think so!
From how he said "a Calc 2 student would try..." I kept thinking the trig substitution was going to hit a wall and he'd have to use Feynman's technique
4 minutes!
A hard integral!
So impressed!
Thanks so much for watching!
@@BriTheMathGuy thank you as well!
Pffft
I can do that in l less than two
@@prajodcp7145 no
@@prajodcp7145 clownin
I loved this. No bizarre tricks, just do whatever the problem is screaming at you.
Thank you.
Really glad you enjoyed it!
Shorter method :
We know integral fx from a to b is same as integral f(a+b-x) from a to b
once we got I=ln(1+Tanx)
I=ln(1+(tan(45-x)) .. expand tan(a-b)=(tana-tanb)/1+tanAtanB
=Ln(1+(1-tanx)/1+Tanx)
=Ln(2/1+Tanx)
I=ln(2)-I
I=1/2ln(2)
Now substitute limits
I= (1/2)ln(2)*(π/4-0)
Done
nice usage of King's property. I like that
You're a JEE Aspirant, aren't you? 😂
Amazing presentation. Done like a real Math Prof!
2:42 In JEE questions, to solve sinx + cosx we usually divide and multiply by √2 to get √2 sin(x+π/4) (common in SHM) it's same but easier.
thats exactly what i was thinking
That trick is widely used in Suneung too
suiiii one of my peeps are here
I knew it I will find someone and see! i did xD
like a criminal undercover
I didn't really find a way to solve this integral without using advance techniques, and you've just showed you're solution like a champ. I'm shookt. Here's a quick sub for ya!
Literally man, you made this question super tough. Just apply ∫f(x)=∫f(a+b-x) for limits a to b. Apply formula of tan(A-B) for A=π/4 and B=0. Add both equations and apply ln(a)+ln(b)=ln(ab). Only ln2 will left. Answer would come (π/8)ln2. Seriously this is a question from one of the easiest books of Indian Mathematics, NCERT.
Your videos are amazing ! Calculus is my favourite topic from Math. I wish I'd seen them while I was still in school. You're a big inspiration
Wow, thank you!
@@BriTheMathGuy no, thank ❤you for these amazing videos !
I got… so excited when I saw that the integrals would cancel out
Awesome video! Thanks for sharing, Bri!
You bet! Thanks for watching!
Johnny Sins of the math world, thanks! I love this video.
😂😂
*Johnny Sinx lol
@@Surajsharma-sj9gx Johnny Sins of the math world = Johnny Sinx it is. Let's make this viral.
@@malexmartinez4007 together we can 😂
@@Surajsharma-sj9gx * ln(e) sinx
Wow! Thanks a lot for the awesome trick with sine and cosine addition!
wonderful explanation bruh!
I have seen lot of your math videos ,It actually helped a lot. :)
I am in my second semester currently , you are far better than my math prof. THANKS A LOT KEEP IT UP
Fantastic explanation!
That was pretty amazing. Great video composition, as always
Thank you very much!
i’m loving your videos! super well put together! great content, great editing, and you explain things really well, you’ve got yourself a new subscriber
Thanks so much!
this was really good. keep it up!
Thank you watching!
Thanks YT for recommending this channel! It’s pure quality
Smooth like butter like criminal undercover
You are doing a good job. Thank you
So nice of you!
This way of solving is so genius
A very elegant and analytic approach! However, I prefer Feynman's trick in this case.
Anyway: I love your condensed and concise way to present the solution.
Easier than Feynman’s trick, and also shorter ! Great
Glad you enjoyed it!
at 1:05 you can use the king property with ln(1+tan(pi/4-theta)) and use the tan(a-b) formula
Thanks for making and sharing another phenomenal video! I was curious: did you know that you needed to end up with the two equal and opposite integrals near the end of the problem, or was that something you stumbled onto or found in a solution writeup somewhere? Either way I think that's amazing; I was just curious about the thought process behind it and how this could help an overall approach to integrals like this. Thanks!
Another method: when we are at ln(1+tan(a)) [let a = theta], we can use king's rule to get ln(1+tan(pi/4 - a)). Next, we can use the tan(a-b) identity, and it will turn out to be ln(2/1+tan(a)). Next, we use the quotient property of ln to separate the integrals and turns out we get the original integral. therefore, assuming the original integral to be I, we get 2I = pi/4*ln2, which evaluates to pi/8*ln2.
A simpler way to do it is to use the substitution x=1-t/1+t
The integrals have gotten cooler.
beautiful content as always
I appreciate that!!
That's really cool! Thanks BriTheMathGuy :)
there is a much simpler way to solve it. When you have the ln(tan(theta)+1)) just use king property to get ln(tan(pi/4-theta)+1) and it just simplify like butter.
You can also try x = (1-t)/(1+t) it work good with log.
Thanks for explaining the sin x + cos x identity! This is my new favorite formula now... why didn't they teach it in schools?
(no I am not a JEE aspirant)
They did for me (in high school)
Definitely had to make use of that formula in differential equations.
They did even for me. (Jee)
Didn't learn that specific identity - but we did learn the method (harmonic form)
In india they do
If u r not fm india
Then howdo u know abt jee
I started off with by parts and understood that its gonna take me nowhere
😅
You could have just used the identity that integral from A to B of f(x)dx = integral from A to B of f(A+B-x)dx .
Also works!
Wait that’s a thing?
@@ridlr9299 just use u substitution and you'll see it works.
I see no reason to apply that here, as you will make the function even worse,
Or can you elaborate on that please..
@@rishu_Kumar07 I meant on the log(√2cos(pi/4-x)
In 1:07 instead of writing tan as sin/cos we can substitute theta for pi/4 - theta and we get the answer in 2 more steps
I actually just came across this in another problem I was solving!
Very cool!
It will be much more easier
Put I =lim 0 to π/4 ln (1+tan theta)
= lim 0 to π/4 ln{1 + tan (π/4- theta) }
Next put the formula of tan (π/4-theta)
Then I = ln (2/(1+ tan theta))
2I = lim 0 to π/4 (ln 2)
And thus I = ln 2 by π/8
kings rule?
Yep I did exactly the same when I first saw this problem
Brilliant job. You showed it so clearly.
Thank you! Cheers!
This problem was a regular problem in our textbook. Another easier way would be just to use proven property of a definite integral. But got a new way to solve. Thanks! This was interesting ☺️
Great tip!
This was so helpful!
Really glad you thought so!
Oh why couldn’t this have realised earlier when I had to do this question for my Australian hsc maths ext2 assignment. Just kidding great video, alternate format which I used however was to use the law that you used at the end of the video when you have the integral of ln(tan(theta)-Pi/4), where you can use the tan compound angle formula and do some simplifying in order to get 2x your original integral as equal to pixln(2)/4, giving the same answer as you showed here
You can also use king rule for solving ln cos terms
I love your editing style brithemathguy
Thanks so much!
very well done indeed.
Thanks! Have a great day!
Video title is fantastic . 💖💖
There is also a easy way to solve this by taking
TanΦ as tan(π/4 -Φ)
Crazy that how i got this same integration problem for my finals in grade 12
I'm in highschool and I have no idea what have I just watched.
Maths is designed that way
Wew lad solving that integral was a thrill ride
Waiting for u to get Silver Play Button 🤩 You deserve it man….
Thanks so much!!
@@BriTheMathGuy 😍🥰
2:55 animation gives me chills for whatever reason. Great video!! ☀️
Thank you!! 😁
just a recommendation: you'd love the channel 3blue1brown then, probably you already know about it
@@nikhilnagaria2672 I think Bri uses his library for visualizations
After substituting x=tan(theta), Substitute theta=pi/4-T. To rewrite the integral then use the tan(A-B) identity then use the log properties to see the magic happen you would right away get your answere
Amazing!
Glad you thought so! Thanks for watching!
The hardest butter you have ever sliced in your life
Well, it literally looks really similar to one of the questions in HKDSE 2020 Maths Exam (The HKDSE examination is the university entrance examination for all high school students in Hong Kong)
I ran into a question like this yesterday, except it also involved partial fraction decomp, and when the terms started canceling out like that I though I did something wrong. And went about it a little different. Now I’m gunna have to go back and look to see if I would have gotten the same answer.
1:06, just use king's property after that and add the integrals, get the answer in 2 lines
The change of variable u=(1-x)/(1+x) leads to solve easily this integral.
The presentation just amazing
Can you make videos about what really single,double and triple integration shows?
These videos need more comments!
Really slick integral 👍
Glad you thought so!
@@BriTheMathGuy for a minute I thought I had fat fingered and missed the L in slick 😊
Him: “you probably want to do a trig sub”
Me :IBP! IBP! IBP! IBP!
new viewer here. takes a while for the compi but gets it done:
*Integrate[Log[x + 1]/(x^2 + 1), {x, 0, 1}]*
I wish I knew calculus so I could understand your integral videos, but I'm still years away
You will get there!
Smooth like butter
Like a criminal under cover
Thank you that was great :)
Thank you so much.👌
Some time ago i saw integral
Int(x/sqrt(e^{x}+(x+2)^2),x)
Try to calculate it, and it is not nonelementary as you may think (I can do it , my way involves tricks like adding zero , multiplying by one and then doing obvious substitution)
CAS programs like Mathematica or Maple are unable to calculate integral i gave
I prefer u=(1-x)/(1+x) substitution for your integral
You should have used kings rule after you get integral of ln(1+tanx) it's faster
This is also in the "école polytechnique de paris" oral exams (oraux X-ens analyse tome 1 pour les fr)
There also you can do another method by not opening tan into sin and cos and only solve in tan by identity tan a + tanb. Then seperating them and put the limits
Love this integral! I solved this using integration by parts
Nice!
this was one of the problems on the putnam
I did it by setting u=ln(1+x)
How i wish i was college again... This video is very helpful..
Glad to hear that! Have a wonderful day!
It's like you're the Bob Ross of mathematics
The solution for x^2 = 2 is +- root 2 and not just the positive. If the answer was carried out you'd end up with two answers, being the positive and negative version of the one you just got.
This kind of problem solving makes me wonder how I survived 1st yr uni maths…
i love how math class is just learning greek
For the cos(a+b) you could Just segue that sin=cos because of the Same coefficent. Grest video
Eh, this integral had pretty much everything I hated about Calc II. And I loved i and iii.
what was the little coefficient trick at 2:30? I don't think I was ever taught that in school
like a criminal, undercover
You should also show a plot of that function on that interval 😉
Good idea!
@@BriTheMathGuy just looked for this in Wolfram Alpha and it doesn't match www.wolframalpha.com/input/?i=int+from+0+to+1+of+ln%28x%2B1%29%2F%28x%5E2%2B1%29
@@joeeeee8738 What do you mean, Wolfram gives the exact same answer
@@flyingpenandpaper6119 yeah, my bad!
Smooth like butter, like a criminal undercover!
Compute the Integral x^2024/(x^2+1)^2023 dx
2 min integral for all jee aspirants
Why wouldn't you use e to remove the natural logs and then put the natual logs back in after
That was sick
As a jee advanced aspirant, this problem seems so easy!
Substitution x = (1-u)/(1+u) will work
nice!
Why didn't we take the rule i.e∫ 0 to a f(x) =f(a-x) in the step of.∫ 0 to a ln(1+tanΘ) please clarify
Good one.
Thank you! Cheers!
4:32 surely you mean "antidifferentiate."
both are same.
What about the integral of cos(x)/(1+x^2) from 0 to 1?
Used king property for that one