an A5 Putnam Exam integral for calc 2 students

Integral of ln(x+1)/(x^2+1) from 0 to 1, this integral was on the Putnam exam back in 2005. We will use a trig substitution and some integral properties to take care of this problem!
This integral is also called the Serret's Integral. Subscribe to ‪@blackpenredpen‬ for more fun math videos.
Check out Mu Prime's video on the integral of ln(x^2+1)/(x+1) from 0 to 1, • It took me 3 hours to ...

[@NadiehFan]

Historical note: this integral was first discussed by Bertrand in 1843 in the Journal de mathématiques pures et appliquées. Bertrand used the technique which is now often incorrectly attributed to Feynman by American sources. In the next issue of the same journal Serret solved the integral in a single line, which is why it is also sometimes referred to as Serret's integral. Bertrand believed his technique was new, but in fact Euler had already discussed introducing an additional variable and differentiating under the integral sign in his 1775 paper Nova methodus quantitates integrales determinandi (E464 in the Eneström index). The title of his article indicates that Euler also believed the technique was new when he wrote his article. But Leibniz already used the technique in an appendix of a letter to Johann Bernoulli dated August 3rd 1697, so it is older than Euler and indeed it is therefore also known as the Leibniz integral rule. Clearly the technique dates back to the very beginnings of calculus as we know it, so there is no justification whatsoever to attribute this technique to Feynman. For the record, Feynman never claimed that he discovered the rule. He learned it from an old calculus text that his high school physics teacher had given him. *References* (replace every + with .) portail+mathdoc+fr/JMPA/PDF/JMPA_1843_1_8_A7_0.pdf portail+mathdoc+fr/JMPA/PDF/JMPA_1844_1_9_A41_0.pdf scholarlycommons+pacific+edu/cgi/viewcontent+cgi?article=1463&context=euler-works archive+org/details/leibnizensmathe00leibgoog/page/n458/mode/1up?view=theater hsm+stackexchange+com/questions/8132/why-is-differentiation-under-the-integral-sign-named-the-leibniz-rule en+wikipedia+org/wiki/Leibniz_integral_rule

[@prydin]

I'm not in college, I rarely have to use integrals, yet I make sure not to miss a single episode of this channel. Excellent brain gymnastics. And if I'm ever stuck in a deep dark forest and need to integrate some obscure function it will become helpful! :)

[@lazzy25803]

4:02 It can be solved really easily if at this moment you apply King's Property and get Integration as ln( tan(π/4 - x) +1 ) which is equal to ln ( (1 - tanx)/(1+tanx) +1 ) = ln( 2/(1 + tanx) ) so we get I = Integral of ln(2) from 0 to π/4 - I hence 2I = ln(2)π/4

[@ravikotadiya3403]

Yeah bro I also think this method 👍👍👍👍

[@clevertech8342]

Yes property 4 yeah bro true

[@kushaankoul5775]

literally a standard integral atp pi/8 log 2

[@marcopicione3432]

Warning!!! For your mental safety do not try integration by parts

[@hamzahhussain6580]

Luckily this is not in A level Mathematics. Integration is easy in A level, the hardest it probably gets is repeated integration by parts

[@TheJampt]

By parts it is not hard. Try integrate by parts twice and you get a syllogism 0=0.

[@anymeskywalker5463]

I used and got the answer

[@SuperSilver316]

yup it doesn't give you an answer in terms of known functions, no matter which way you take it. You part of the final answer on the first by parts, which gives an interesting result for another integral, but you wouldn't know that without doing the way he did it in the video.

[@Jamie-ef9dl]

TheJampt can you please explain how it’s done by parts cause I get massively stuck when I have two of the same integral and can’t just add it to both sides or smthn

[@sergiokorochinsky49]

Flam's parametric method is very elegant. BPRP trig substitution was clever and inspired. I just solved it by brute force, but got some interesting results worth sharing... First I replaced the whole integrand by its series expansion Log[x+1]/(1+x^2)= Sum[a_n x^n,{n,0,inf}] Then the integral is just Sum[a_n/(n+1) x^(n+1),{n,0,inf}] Common sense indicates that the series should be centered in 0 or 1, so half of the terms disappear, but the fastest convergent solution was with the series centered in 1/2, as the coefficients get divided by powers of 2. For the series centered in 0 (McLaurin) the convergence is extremely slow, as the coefficients converge to Log(2)/2 and pi/4 alternating signs. Curiously, the solution is the product of these 2 convergence points. Of course, all these trials only gave a numerical approximation. Then I tried to replace just the log for its series expansion, and integrate each term by doing long division. The integrals separate in 2 groups: log(2) and ArcTan(1), each one with a finite number of additional terms making a triangle. The funny thing is that both triangles complement each other, giving the series expansion for log(2)/2 and ArcTan(1), cancelling each other and giving the analytical solution!

[@sebmata135]

This integral and the other Log integral by symmetry are amazing. You never fail to blow my mind BPRP!

[@blackpenredpen]

sebmata thank you!!!!!

[@u.v.s.5583]

The whole solution is nothing short of miraculous, but once you wrote down the integral of log(cos(theta-pi/4)), the rest was completely trivial and waste of time. I mean, the limits are from zero to pi/4. You need no further explanation to cancel the log cos beasts.

[@surajkulkarni565]

Here is one more fantastic solution kzbin.info/www/bejne/q2SrpXZ-n8-GgNU

[@TheRedfire21]

Thats so clever, I tried finding an antiderivative but is just way too hard and not feasible if you are not a machine. sometimes I forget definite integrals are just like calculating areas and you can clearly see that the area of the cos(x) function and the cos(x-pi/4) is the same on the interval [0,pi/4]. thanks for the video!

[@blackpenredpen]

My pleasure!

[@hemantpandey7539]

It works in general too! When integrating f(x) from a to b we can also integrate f(a+b-x) from a to b to get the same answer. It's kind of like calculating the area from the other side :P

[@mike4ty4]

The antiderivative can be solved with the de Jonquiere's function, also called the polylogarithm: Li_s(x) = sum_{n=1..inf} (x^n)/(n^s) which is simple generalization of the logarithmic Mercator series: -ln(1 - x) = sum_{n=1...inf} (x^n)/n. In particular, one should note that int Li_s(x)/x dx = Li_(s+1)(x) + C. One should become familiar with that there are a lot more different mathematical functions in the toolchest than just sin, cos, tan, exp, and log, especially now with computers and math libraries to calculate them readily at hand. I would suggest BlackPenRedPen illustrate this. sin/cos/tan/exp/log works fine for Intro Calc 1 and 2 but when you get on up to stuff like Complex Analysis or out into the real world, you'll want to have a lot more in your bag with you.

[@pranavsingla5902]

Learn integration then bruh...

[@aashsyed1277]

@@blackpenredpen m6 0l3àß743¡ It's m6¡

[@FuhrerShattercore]

You solved what took my friends and me a day in 19 mins

[@fourier07able]

This integral is not so easy .

[@pi17]

I solved under 5 minutes lol.... We don't have so much time in exam

[@abhinavtushar305]

A day lol this took me only 30 seconds,just substitute x=tanx,and use the king's rule

[@pi17]

@@abhinavtushar305 there's no king rule.... It's just a property........ Bansal sir used to call it that..... It's not an internationally used term.....

[@abhinavtushar305]

@@pi17 😅😅

[@karansgarg]

I honestly did not see that coming, that was incredible

[@Kyle_da_athlete]

This was an absolutely gorgeous integral. I got up to splitting it into the 3 separate integrals but I didn't think of working it as area at once. I wanted to get an indefinite integral and them substitute. Thanks for opening my mind to this way of thinking

[@ivornworrell]

*This Vietnamese brother knows his University Maths but @**03:35** when you wrote the limits of integration from 0-1 I was wondering "What the hell is he doing?" but when u corrected urself @**04:15** & I said "Ah yea, he's back on track, 0-pi/4 is correct" ! Good stuff.*

[@shajiruby2346]

Very convoluted explanation. Using property of definite integral, the problem can be solved in less than a minute

[@mrjnutube]

Never ceases to amaze me. Awesome...

[@blackpenredpen]

Thank you!!!!!

[@Philgob]

i mean... i understand it but i would never ever have came up with that on my own

[@furiousfajitaa2367]

That basically sums up all maths haha

[@Saki630]

I dont know how I was recommended this video considering I have probably 70% of my subscribed channels never being recommended to me..... This is a good video for students learning calculus.

[@lordkitsune2713]

It's known as the Serret’s integral for some people who are still wondering.

[@blackpenredpen]

Oh nice!! I didn't know about this.

[@Rangertom6]

There is actually a very lovely way of doing differentiating under the integral sign for this integral as well! if you let I(a) = integral of (ln(ax+1)/(x^2+1)) dx you can also get the answer!

[@djsmeguk]

this I would love to see!

[@ishaangupta2185]

This is how I originally solved the integral - Feynman's technique! I was curious to see how it could be solved using only calculus 2 techniques.

[@tfg601]

@@ishaangupta2185 Hmm I did Feynman's technique up to a certain point and I couldn't figure it out after I differentiated I tried usub and it didn't work, could you please explaiin how you did it? Thanks if you do

[@ishaangupta2185]

@@tfg601 It was months ago but if I remember correctly, it was to set P(t) = integral of ln(1 + tx) / (x^2 + 1). Then differentiate and solve like normal.

[@tfg601]

@@ishaangupta2185 Bro I differentiated with respect to t and even check on wolfram alpha that is not integrable with respect to x

[@injanju]

Can you please do another Putnam integral, the integral of dx/(1+(tan x)^sqrt2) from 0 to pi/2 ? I'm really amazed at how you come up with these clever solutions, really mind blowing!

[@hamiltonianpathondodecahed5236]

Wolfram alpha

[@sarangisdon]

Apply King's identity for definite integration

[@pranavsingla5902]

Use that the integral remains same by putting x as (pi/2 - x) and add both of them. The numerator and denominator cancel out....the answer is pi/4.☺☺

[@AdityaKumar-gv4dj]

4:00 you could have used King's Rule at this step, ln(1+tan theta) will get cancelled after adding I's to get a 2I. We get 2I= pi/4 ln2. Therefore, I=pi/8 ln2

[@Arthur0000100]

I solved this exact integral just the day before yesterday! Noticing the tan^2(x)+1=sec^2(x) for tangents is half the work. Keep up the good work!

[@vector2817]

The first substitution was really clever, I like how you could then solve the rest of the integral using the knowledge from previous integrals ;)

[@adonaythegreat8426]

All your works are clear and easy to undestand. Good on you.

[@Anas-nu7io]

Great video! I love these interesting integrals. Keep uploading my friend!

[@jihanhamdan5465]

Had two awesome ways to solve it, thanks to u and to flammable math

[@sergioh5515]

Awesome integral!

[@sergiokorochinsky49]

Integrating by parts, it's very easy to prove that the integral between 0 and 1 of ArcTan[x]/(1+x) it's also pi/8*Log[2]

[@blackpenredpen]

Sergio Korochinsky it's very easy

[@sergiokorochinsky49]

two integrals for the price of one...

[@kne-si5zj]

how would you prove that

[@sergiokorochinsky49]

@@kne-si5zj ...just integrate by parts. dv = 1/(1+x) then v = Ln[1+x] u = ArcTan[x] then du = 1/(1+x^2) u v = ArcTan[x] Ln[1+x] (between 0 and 1) = pi/4*Log[2] and the integral of v du is solved in the video... pi/8*Log[2].

[@souleater9189]

whoa whoa whoa... i thought this was black pen red pen, not black pen red pen blue pen!!

[@blackpenredpen]

hehehehe

[@thecation839]

lol

[@rahul7090]

There's much easier way to do this. After you reached I=integral of log(1+tanx)dx with limits 0 to pie/4. Let's call it first equation.I then straight way used f(a)=f(a-x) property. Then I =(2/1+tanx)dx with limits 0 to pie /4. We call it second equation. Adding I) and ii), we get 2I=Integral of (log 2)dx with limits 0 to pie/4. Since log 2 is constant , we take that out of integral and bring 2 from LHS to RHS. Then I=1/2* log2 integral of xdx with limits 0 to pie/4 and we get same answer.

[@n0ita]

THAT'S GENIUS !!!! In the final part i had a mindblow.

[@hemaharish]

There is another easy way to solve this. After the theta substitution once we get the integral as integral from 0 to pi/4 ln(1+tanx) dx (using x for theta), we can use a property of definite integral that integral 0 to a f(x)dx is the same as integral 0 to a f(a-x)dx. So the 0 to pi/4 ln(1+tanx) (let us call it capital I), is the same as 0 to pi/4 ln(1+tan(pi/4 -x)). If you use the trigonometric formula for tan(pi/4 - x) and simplify the integral becomes integral 0 to pi/4 ln(2/(1+tanx)) which is integral 0 to pi/4 ln2 - 0 to pi/4 ln(1+tanx). This means I = 0 to pi/4 ln2 - I, means 2I = Integral of 0 to pi/4 ln2 means I = pi/8 times ln2.

[@qiaozhou9296]

Thank u for your patience with each little step!!!!!

[@ashuprakash6266]

A video to make people feel good about Putnam exam. This is like ABC of what's asked in their

[@oratval]

at the beginning of the year I was taking calc because I had to. Now I'm spending my free time watching videos of people solving calc

[@luohuo2001]

He makes calculus so easy, if only he was actually my home tutor

[@reaper3.097]

our* home tutor

[@Zaytoven21]

Very Niceeeeee i think i would like to see more of these Math competition problems , as this had a very satisfying way of solving it .

[@blackpenredpen]

Thanks!!!!

[@YASSIRMESSIooo]

ؤ

[@Byt3me21]

A graph would have been icing on the cake. Thank you.

[@fourier07able]

What an ingenious tricks! Firstly the change of variable x = tanθ , secondlysinθ + cos θ = √ (2) cos( θ - π/4), thirdly making use of cosine is even, i.e. cos (- Ø ) = cos (Ø ), plus other minor tricks. Impressive! This integral is one hardest out of integrals solved on terms of elementary functions I ever seen.

[@jeffreyluciana8711]

Whoever was the first person to figure that out should get a star for their forehead

[@sss-ol3dl]

nice! i managed to do it by differentiation under the integral with ln(ax+1)/(x^2+1)

[@hansenchen1]

Very nice!

[@algorithmvideos]

Please provide me the solution if u can by your method

[@heartpirates245]

I was gonna say, I wonder if this is possible with Feynmans technique, looks like it is

[@cpotisch]

@@algorithmvideos Two years late, but the process is: partial differentiate, do partial fractions, integrate in the x world, integrate in the b (parameter) world, eliminate the constant by evaluating at I(b=0), then relate the value of I(b=1) to itself and solve.

[@algorithmvideos]

@@cpotisch thanks for the same.. U took your time..😁😁

[@ismaelcastillo188]

I used Leibniz rule introducing ln(tx+1) and it worked marvelously

[@donmoore7785]

Awesome. That trig identity for sin theta + cos theta rings a very faint bell. Well done.

[@HasXXXInCrocs]

If I hadn't seen this video and this we're on a test I'd probably drop that class lmao. Amazing work, crazy how powerful substitution is. Hope I can apply this knowledge one day!

[@gatoradeee]

Beautiful and elegant. More putnam problems please.

[@maximedelboo9021]

Happy to see some really hard integrals returning, they had been gone for a while... please do more putnam!

[@blackpenredpen]

Thanks! I will try.

[@rubikscuber1114]

Ohh my god this question came in my today's exam.The exact same question THANK YOU SIR!

[@suryatallavarjula3184]

What kind of exam?

[@MiroslavMakaveli]

Very motivating integral. Thanks a lot.

[@blackpenredpen]

Miroslav Dimitrov my pleasure!!

[@todayscalm]

Thank you so much! I am a big fan of yours. That's it!! ^^

[@EAtheatreguy]

I did this with Feynman's technique, I(b) = int [0,1] (ln(bx+1)/(x^2+1))dx, but it only worked because the endpoints were the same as the parameters you need to take the integral to 0 and to equal the original integral.

[@user-wx8bm1pg1d]

I was just about to give up until I saw this comment and finally noticed the thing about the endpoints

[@Supernova799]

It is a beautiful integral

[@user-vm6qx2tu3j]

Math is love, math is life🙌

[@anesumurwira8453]

You are a real teacher, I like it. Thanks a lot, it helped me preparing for my final G17 exam.

[@a5hen]

when we get integral of ln(tan theta +1),we can simply use the definite integral property of f(x)=f(a-x) when f(x) is integrated from 0 to a.We then get ln(1+tan(45-theta)).After expanding tan(45-theta) and simplifying the expression inside the integral,we are just left with ln(2).So this could be done in a much simpler way. Anway,great video.

[@isaquepim4555]

I'm in school vacation here in my country(Brazil), and I should have been resting, but I just can't stop watching these integral videos. Nice job BPRP 😁

[@rashmigupta6227]

Simple: Do the substitution x=Tan§ And then after obtaining the expression just use King rule and just add both , and we have achieved our goal

[@qedmath1729]

For those who don't know the king's rule, it states that If you have an integral with bounds a and b f(x)dx its the same as integral bounds a and b f(a+b-x)dx.

[@messinios97]

Happy new year BlackpenRedpen. That was really nice integral, it was awesome!!! Greetings from Greece!

[@oguzhanyesildemir5119]

Man this was good thanks . I have been trying to solve this integral for hours .

[@thegreencan1286]

I feel like very few people would looks at this problem and try to apply high school methods - the integrand very clearly doesn’t have an elementary antiderivative. But somehow this method works!

[@DragonLoad49]

You are awesome, and that is a cool mic! Thank you for the practice!

[@crancowan8020]

Pretty cool. I did it another way by starting off with u = tan^-1 (x). The integral(s) then become easy to solve using the power series for ln(1+x). You just get an infinite amount of them which when displayed in matrix form (at x = 1) you can see that the result comes down to -2\ln\big(\cos \frac{\pi}{4} \big) +\ln\big(\cos \frac{\pi}{4} \big) = \frac{\pi}{8} \ln 2

[@sriniramiah6445]

Hi Blackpenredpen, Thanks for the great math videos. I think this problem seems too easy for putnam. Your solution is very nice . You can also do it as: STEP 1: Let I = Lmt 0 .. 1 Integral Ln (1+x)/(1+x^2). STEP 2: Let x = tan a. So I = Lmt 0 ... pi/4 Integral Ln (1 + tan a) da. STEP 3: This can also be written as: I = Lmt 0 ... pi/4 Integral Ln (1 + tan (pi/4-a)) da. STEP 4: which simplifies to: I = Lmt 0..pi/4 Integral Ln ( 1 + ( 1 - tan a) / (1 + tan a) ) da. STEP 5: I = Lmt 0 .. pi/4 Integral Ln ( 2 /( 1 + tan a ) ) da. STEP 6: I ={ Lmt 0 .. pi/4 Integral Ln(2) da} - { Lmt 0 .. pi/4 Integral Ln(1 + tan a) da } (the second part is just the same integral I, in STEP 2 above) STEP 7: so we get, I = Lmt 0..pi/4 Integral Ln(2) da - I STEP 8: Move the I to the left side and applying the limits on the right side we get: 2I = pi/4 * Ln2. STEP 8: ==> I = pi/8 * Ln2.

[@kshitijgaur9635]

very nice, soon you'll have 100K subscribers. Great job!

[@mrniemathschannel]

This trig substitution approach is veryinteresting. However，if u=arctan(x) and v = ln(x+1) are recognized, then it immediately follows that the antidrivative is arctan(x)ln(x+1) - ∫1/((x+1)(x^2+1))dx. The second integral can be calculated using partial fractions.

[@svk2542]

It can be done very easily by using properties of definite integral

[@YogeshKumar-rf5ys]

You can use king's property for solving integral of ln(tanx +1) from 0 to pie/4

[@animeepisode9280]

Man I solved it in 1min it boosted my confidence man. Now I would focus on physics portion for my exam.

[@colinpountney333]

Another approach is to expand ln(1 + tan(theta)) as a power series and integrate term by term. You end up having to combine two infinite triangular arrays but there is a neat trick and the answer drops out easily once you spot it. You need to use the identities 1-1/2+1/3-1/4+......= ln(2) and 1-1/3+1/5-1/7...= pi/4.

[@veled806]

you can also solve this problem with x=arctanu substitution and then applying king's property, adding the two integrals you got before and you get the same answer.

[@zeldasama]

You always have a tendency to talk fast. You clearly love doing math, and I love that, keep up the good work.

[@tomatrix7525]

Aha what a legend! How do you actually think so cleverly? When I watch this it makes sense but its so clever man!!! Keep it up I marvel at your work!

[@adhirachannel1148]

When u bring integral in thetha world i.e. integral ln( 1 + tan(theta)) then use kings property n get answer in half a minute

[@aaishikdas]

Good idea bro😯😯....that's great🙂🙂

[@jeremybuchanan4759]

Very nice - I thought there was no hope I would understand as you started the second substitution but I made it through!

[@jeffreyh.1436]

hmm this Putnam problem seems strangely similar to SMT 2019 Calc #10... that being said, here is another (cleaner imo) way based on the SMT solution substitute u=x+1 to get int_1^2 ln u/(u^2-2u+2) du we want similar integral with same bounds, so we substitute v=2/u to get that I=int_1^2 (ln 2-ln v)/(v^2-2v+2) dv adding the two gives 2I=int_1^2 ln 2/(w^2-2w+2) dw trig sub gives the desired pi/8*ln 2

[@MrExtremehustler]

That was so satisfying.

[@deepambanerjee5594]

It will become easier if you apply Kings rule in log(1+tan theta)

[@verainsardana]

yes i was about to comment that

[@gaunterodimm7219]

yeah we do that in our +2

[@SumitSharma-wf5xb]

Sahi khel gaye😀😀😀

[@soap6448]

He’s writing with two markers..... what a teacher

[@prollysine]

Hi bprp, wonderful number game with angle functions...

[@blackpenredpen]

Thank you 👍

[@petegentles4211]

your videos are addictive.

[@harley6659]

That was so... Elegant

[@sardarbekomurbekov1030]

Interesting integral and interesting solution.

[@Squeaks99]

Sir just awesome

[@jjwesker1061]

Well... this IS a potential AP Calculus Question (I take AB). I’m glad I found you because I can keep my Calculus strong during Spring Break! Now I will just see more videos to know how to do harder integrals.

[@mihaiciorobitca5287]

Ooo !what elegant is bprp and also his equation !

[@amanmahendroo1784]

who else thinks this is a little too easy for Putnam?

[@debdhritiroy6868]

yeah man, I already did this in school... So way tooooo eady for Putnam... But, if true, I can maybe then get rid of my fear of Putnam

[@saikat93ify]

I think, it's become a famous question after it was asked in Putnam. But at the time, it might have been a very new question. But yes, it is indeed very easy for Putnam.

[@gurkirat2001]

Yes A easy one

[@chandrabhan7212]

Yes, easy because it's the "standard" question when you're introducing students to Leibniz rule. Quite sure back then it would've stumped most of us (it's Putnam after all).

[@osamafawzy4164]

@@chandrabhan7212 jþt

[@saikat93ify]

Fantastic explanation !

[@wasiqmk]

It's is "Serrets integral" Tan substitution And properties of integral will help you crack this integral π/8log(2)

[@thel0negh0st]

another method is by writing the integral again substituting theta in ln(1+tan{theta}) as (pi/4-{theta}) Then by adding both integrals you can reach to the same answer in next two/three steps.

[@mr.soundguy968]

Feynman's trick with ln(tx+1)/(x^2+1) also works

[@AndDiracisHisProphet]

Ok, because it's the new year and you did a great job (as always) extra for you I unsubscribe, so that I can subscribe again

[@blackpenredpen]

HAHAHAHAHAHAHA!! Thank you. I wish all my subscribers can do the same for me! = DDDDDD

[@ivornworrell]

AndD whats the point of subtracting then adding back the same quantity since the original quantity will remain unaltered? well thats exactly what ur doing here AndD, by unsub. then subscriging again!

[@felipealonso7859]

@@ivornworrell Obvious. It's like that integral you have to add 1 and then subtract 1 and you get a nicer little integral...

[@anticorncob6]

Do 2017 Putnam A3! I was able to prove that it approaches infinity but not that it is strictly increasing.

[@byounger9171]

Currently watching as an AP Calc 1 student in highschool and let’s just say I’m scared for my life after seeing all of this

[@sergeirachmaninoff6397]

This is as hard as it gets. My college wouldn't have this on a test bc that trick in the 5th line with the sum of sin and cos of theta is considered just pointless memorization for them. But there's a probability that you'll find something like this in college, but it isn't likely that this type of problem appears on your hs tests

[@eric336]

Interesting how in all craziness the result turns out to be that simple.

[@lunstee]

My solution was largely the same as yours, but was a little cleaner in two ways. First, when seeing tan(t)+1, my immediate reaction was to substitute 1=tan(pi/4) so that all terms take similar form which facilities combining them. It's easy to show tan(a)+tan(b)=sin(a+b)/cos(a)/cos(b), and using this, the integrand becomes simply ln(sin(t+pi/4))-ln(cos(t))-ln(cos(pi/4)) Which is essentially what you work with on the right side of the board, but written in terms of the sin rather than your cos. You make the equivalent substitution at the bottom left where you juggle with fitting A and alpha=pi/4, but identifying this alpha earlier lets the A take care of itself. From this point, rather than working algebraically, it's easier to look at the first two terms graphically. Just plot sin(u) and shade in the area from pi/4 to pi/2 - this is the range of sin considered in the first interval. Similarly, if you plot cos(u) and shade in from 0 to pi/4, you'll notice the shaded area is simply the mirror image of that from the sin. Taking ln(..) of these is just identical vertical distortions of both, giving again identical areas. Hence the first two integrals cancel each other. This leaves just the constant term which if we multiply by the pi/4 interval gives the final result.

[@ommohamedommohamed9685]

Brilliant job

[@1103juancho]

OMG congratulations , you are really amazing

[@Maxman013_]

Ok that was actually amazing

[@mathematicsmi]

Nice evaluation..

[@ayoubelgoubi9691]

Dude, could you have made this any more complicated, I screamed multiple throughout the proof! Thanks for the vids BTW.

[@punyagupta3071]

We don't need polar form. After ln (tan theta + 1) we can use king rule which gives ln (2/1+tan theta) Finally we get 2I = integration of ln 2 going from 0 to pie/4

[@warrengibson7898]

Always fun to see pi pop up with no circles in sight.

[@juno7424]

This is great! I’ve only take AP Calculus AB before and wanted to see how much I’d get (obviously barely anything), I’m taking calc 2 freshman year in university and wanted to make myself less scared