A Cambridge Integral Experience

[@dombacino7669]

“Well why, because it’s on my shirt” no further explanation needed

[@yusefzaman5242]

QED

[@MrUtah1]

Source: shirt

[@m1n3c4rt]

proof by shirt

[@abubakar-serdiq]

Mentor.... I have been following your Lectures for the past 1yr. I have benefited alot from you sir.. keep the Good Job. Now I have decided to come up with a channel which that I will be teaching mathematics in my native language, to make Maths easier for my society members. Your students from Nigeria

[@remzidag8735]

Vincent Abubakar players in Beşiktaş...

[@mujtabanadeem3901]

Respect++

[@mcalkis5771]

Best of luck to you.

[@abhirupkundu2778]

​@@mcalkis5771all the best*

[@toby.barnett]

This is amazing!! Please do more STEP questions - they are lots of fun and your videos would be super helpful to people sitting it not only this year but in years to come! :D

[@finnhogan5525]

You're analogy of step questions being like a having a tasting menu was really funny and creative. I have been doing the step past papers for over 2 years now and I have never thought of it like that.

[@blackpenredpen]

Thanks!

[@anandk9220]

@@blackpenredpen Can you solve this annoying question please? I'm trying a lot but have no clue as of now. Integral from -pi/4 to pi/4 of { [ (sin x)^6 ] + [ (cos x)^6 ] } ÷ { 1 + (6^x) } I came across this problem on toanmath.com. Thank you.

[@cycklist]

Love this 👍 Classic BlackPenRedPen returns 😀

[@anshumanagrawal346]

I would never have thought I would be able to understand every step of such a hard-looking problem

[@krishnanavati3715]

Great video - step questions are really fun and lots of interesting ones come up. Hope you do some more!

[@insouciantFox]

Woohoo! I got this one! All by myself! Had me puzzled before I found the trick. Here's how I did it: Let the value of the integral be Q and the integrand be f(x)≡n(x)/d(x). Due to some theorem I can't remember the name of, the integral from a to b of f(x) is identical to the integral of f(b-x) over the same bounds. n(π-x)= 2(π-x)³-3π(π-x)²= 2π³-6π²x+6πx²-2x³ - 3π³+6π²x-3πx² = -π³+3πx²-2x³. Since sin(x)≡sin(π-x), d(x)=d(π-x). Adding the integrals of f(x) and f(π-x) yeilds a value of 2Q. The resultant integrand is: N(x)/D(x) := (n(x)+n(π-x))/d(x). Specifically: N(x) = (2x³-3πx³)+(-π³+3πx³-2x³) = -π³. D(x) = (1+sin(x))². Factor out the constant from the integrand. Then complete the difference of squares by multiplying N(x)=1 and D(x) by (1-sin(x))². This results in N(x) = (1-sin(x))²= 1-2sin(x)+sin²(x) and D(x)=cos⁴(x). Split the integrand into A, B, and C. A = sec⁴(x)dx = sec²(x)(tan²(x)+1) = u² +1 du B = -2sin(x)sec⁴(x) dx = -sec²(x)(sec(x)tan(x)) dx= -2v² dv C = sin²(x)sec⁴(x) dx = sec²(x)tan²(x) dx = u² du Evaluate the simple integrals. Note that on the boundary tan(x) disappears. What remain is: 2Q = -⅔π³(sec³(0)‐sec³(π)). sec(π)= -sec(0)= -1 Thus Q = -⅔π³ ≈-20.67.

[@pickleyt6432]

The rule you were thinking of is called Kings Rule of Integration… nice solution btw

[@tfg601]

@@pickleyt6432 It's doesn't have a name merely due to the fact that anyone that graduates highschool can derive it. Really it's not a rule it's just "logic" in the math world.

[@echohasbinokiller4]

Me at a fancy restaurant Waiter: Now, to finish your second course, we are you going to serve you... I don't know.

[@babajani3569]

Amazing. I would also like to see you give some STEP 3 questions a go. They are quite a lot more difficult and are more beautiful as well. There are some very beautiful one such as proving the irrationality of e etc.

[@Jasomniac]

Wow, I had to prove e was irrational once I entered college, that must be tough for a high school student.

[@parthgupta533]

Sir I literally owe 50% of my math knowledge to you, been following you since 2020 when I was new in tenth grade and didn't even knew calculus existed, but as time passed I saw thousands of questions being solved by you, I finally can say that calculus is obsessed with me 😂

[@DevTech03]

17:16 . Question becomes much simpler by f(x) = f(a+b-x) . We are only left with negative pi cube In the numerator. It's the simple integral of 1/(1+sinx)^2 which comes out to be 4/3. So 2I = - 4/3 pi^3

[@gigachad2419]

Got this Integral Correct with Perfect Logic!!! I'm so Grateful....Thanks bprp for the Video❤🙏

[@mohammadfahrurrozy8082]

Bprp : its a cambridge exam question Me : oh no Bprp : its *question number 6* Me : (flashback to the legend) *oh no....*

[@mahatmaniggandhi2898]

😄😄

[@fivestar5855]

The legendary question number 6 from Australia...

[@mohammadfahrurrozy8082]

@insert username search on youtube "question number 6 math olympiad"

[@alberteinstein3612]

Oh no no no no no no no

[@mohammadfahrurrozy8082]

@@alberteinstein3612 even einstein himself afraid from it

[@tommc1425]

If you do a substitution u=x-π/2, you can cancel almost everything by symmetry, after the cancellation you resubstitute back to x and you're only left with the integral of 1/(1+sin(x))². There are actually a lot of symmetries on the integral, they're just hidden by the bounds. You can skip the identity completely

[@jewel8482]

I managed to do it by multiple substitutions!

[@NoName-kj2vf]

I did this too. There is also another way to integrate 1/(1+sinx)^2 by multiplying numerator and denominator by (1-sinx)^2. You end up getting (1-sinx)^2 / cos^4(x). Expand the numerator and split the terms and it is fairly simple to integrate

[@ismailberbache8780]

Amazing video! I really hope you do some more STEP questions, they’re really fun and it’s very helpful to 2021 step takers!

[@alwayslistening4444]

I appreciate how you contrasted the WolframAlpha result with the more informative expression of the evaluated integral. The intuition of a mathematician is enhanced by such expressions as opposed to a given expression's decimal equivalent.

[@slavinojunepri7648]

The analogy to the six-course meal at the french restaurant falls in place. How magnificent this integral problem!

[@blackpenredpen]

Thanks!

[@seshnarayan7972]

Dear blackpenredpen, I first used the definite integral property integral 0 to a f(x)=integral 0 to a f(a-x). I got the integral of the rhs as integral 0 to pi (-pi cube -2x cube+ 3pix square)/(1+ sinx)square. Since the first integral and this integral are exactly the same I named them both as "I" and added them to get "2I" so that the variable terms in the numerator would cancel out. So the result was "2I"= -pi cube integral 0 to pi dx/(1+sinx) square so I could solve for "I" and so the result would be I=-pi cube /2 integral 0 to pi dx/(1+sinx) square. But I did not know how to solve this integral so I looked it up on Wolfram alpha and it said that the value was 4/3. So I multiplied -picube/2 by 4/3 which is -2pi cube/3

[@BCS-IshtiyakAhmadKhan]

You can write (1+sinx)^-2 =(cosx/2+sinx/2)^-4 then multiply num. and den. by sec^4x/2 after that put tanx/2 =t now you can easily solve

[@seshnarayan7972]

@@BCS-IshtiyakAhmadKhan oh I see. Thanks for the insight

[@varunjagan4402]

Or write sinx/2 as 2tanx/2/(1+ (tanx/2)^2)

[@moinmalik1320]

You are doing a great service to all those who love and want to learn math.

[@MathElite]

Wow such a complicated integral but beautiful result

[@ekisvioleolivaradamos6701]

I also have no idea until I watch your video... You're so brilliant... More power...

[@yoyoezzijr]

in the last part, i did a w sub which will make it w³/3 + w with the bounds being 0 to 1 (tan0 = 0, tan45 = 1), which is just (1/3 + 1) - (0 + 0) so 4/3

[@morischacter1076]

for the last part if you factor out x and let f(x)=1/(1+x)^2, you get Pi/2 and then you do it once again leaving 2x-3pi/(1+sinx)^2. break up the the numerator, and you get to the same final part much easier and quicker.

[@tkc-g8h]

Amazing video! When I did it myself for the first time,I got 1/2∫π^3/(1+sin x)^2 dx ,but I couldn't do anymore. Part1 is so hard to comprehend the first time I watched the video.I'd like to try again!

[@mohammedhaidar3200]

For step three instead of figuring out the identity of Integral[x^3 f(sinx) dx], I just factored out one x and assumed the rest were arcsin(sin(x)) and applied the first identity once, then I factored another x (or one of the arcsin(sinx) powers) and applied it again; I split the integral in two and applied it a third time, I got the same result.

[@tupoiu]

arcsin is only a function to [-pi/2, pi/2] from [-1,1] so we can't do arcsin of x when it gets nearer to pi > 1. For this reason I think your method isn't complete.

[@dirt3554]

Love the STEP content. It's great preparation!

[@ginopaperino2608]

this man is powerful

[@yamikira6512]

I saw the thumbnail, started doing the integral, finally managed to evaluate it, then opened the video and realized... it's a step question.

[@Eric-dt7bt]

I was preparing for STEP two years ago and I still remember this problem... Me: the first two parts are so easy, I got this Me, thirty minutes later: maybe I should prepare for another year

[@reu.mathematicsacademy8566]

Tao of our time 👍 great problems solver..you teach me a lot... very unique blackpenredpen

[@rob876]

to approximate -2π^3 / 3, use π is approximately cube root 31, This gives the answer -62/3 which is -20.67 to 2 decimal places.

[@dmddjack]

I got it correct before watching your video! btw this is quite a nice and elegant solution!

[@blackpenredpen]

👍

[@erenozturk586]

if you want to skip to straight to dessert, sub in u=pi-x, then add your result to the original integral, then after some nice cancellations let t=tanx/2 :)

[@violintegral]

When I got the integral from 0 to inf of -π³(1+t²)/(t+1)⁴ after the weierstrass substitution I was actually drooling all over this delicious integral. And after watching this video, it's interesting to see that bprp did it in a completely different way, using the results of the previous problems to help him. Really cool integral.

[@erenozturk586]

@@violintegral had the exact same experience haha

[@joshhutchinson7031]

Finally step question getting some recognition!!

[@Xerkun]

This is awesome. Thank you for this! Please do more of this kind.

[@benoist13]

Another way is : 1) substitute x=u+pi/2 to get an integral between -pi/2 and pi/2 then use he fact that odd functions integrated symetricaly are zero and finally, substitute cos u = (1-t^2)/(1+t^2) with du = 2dt/(1+t^2) (where t = tan x/2). Note : in France, we do not use sec x and cosec x but 1/sin x and 1/ cos x.

[@chessandmathguy]

this is amazing, thanks for posting!

[@colfrancis9725]

Very good, thank you. Your method for part 3 is good but not obvious - you went back to the subsitution y = Pi - x. Alternative methods exist and seem more obvious. For example, factorising the integrand as x f(Sin(x)) . (Polynomial in x) to give you something on which you can apply integration by parts (with the earlier results from the question). It's not faster but it's NOT SLOWER and it doesn't require much luck or special insight to make good progress.

[@jeeves_wooster]

Using symmetry here is definitely much more obvious as there's a pi within the integral. Integration by parts is generally much more messy even if it isn't longer. When looking at trig integrals one of the first things you go for is symmetry; before ibp.

[@colfrancis9725]

@@jeeves_wooster Hmmm... Maybe. I'm glad you thought about symmetry, you're a better man than I am. I could easily have missed it. I think the main prompt for substituting y = Pi -x came from experience of the earlier questions (see approx 16:30 in the video for Blackpen's motivation) and that is ONLY partial motivation in my opinion. Before putting pen to paper, you can see (or reasonably imagine) a messy expansion of a (Pi-x) to the third power through which we would struggle to guess that we're going to get anything more useful than some way of integrating the x^3 f(Sin(x)) term in the original integral. It's just good fortune that things simplify better than expected (about 19:40 in the video) by taking TWO components from that expansion over to the other side. If you hadn't been lucky you would have been expecting to repeat the same kind of substitution process with the x^2 f(Sin(x)) term in the original integral. Let's be totally honest, I would expect most people to test the substitution idea on the lower order term, x^2 first before trying it with the x^3 term - there is no good reason to guess that you should start with the higher order term (i.m.o.) By comparison, the method of integration by parts (as outlined earleir) is something where we can see good progress being made on every step. You could teach that method to others AND justify the reasoning on each step easily. You said "ibp is generally much more messy..." but I would argue that the student can reasonably be expected to imagine (before putting pen to paper) that it won't be difficult here: After factoring out x.f(Sin(x)) the polynomial we have left is only of order 2, so it's coming down to order 1 after differentiation during ibp. Already at order 1 - there is no risk of the algebra becoming messy.

[@AdrienLegendre]

You did this the hard way! For integrals over a finite interval, transform variable x so interval of integration is centered at 0. For example, u=(x-pi/2)/pi. The result is a sum of integrals of even and odd functions. Integrals of odd functions are zero. This brings you to the last step of your method. This method has far wider application and is simpler than the method you used.

[@nickallbritton3796]

He was just following the steps of the problem as it was given. Solving it another way in the exam would be pointless after doing all that work in the other parts.

[@narrotibi]

It's a hard life in the calculus world! This was brilliant.

[@nolimitderrick4822]

The hardest part of this integral for me was understanding that what I thought is a Greek letter is just how u right cos😂

[@aalsii]

Q-6(B) was asked by IIT for Msc. mathematics exam too.

[@goodra4999]

I'm a Japanese college student. This lecture is very interesting and usuful for studying English, So it is wonderful. I'm sorry in poor English🙇

[@efulmer8675]

Your written English is pretty good! I've met native English speakers with worse English than you so don't feel bad if you mess things up.

[@bishalbhattacharjee5930]

Sir please make a video of complete set theory with relations and functions. I didn't understand set in maths.

[@MartinPerez-oz1nk]

THANKS PROFESOR!!!, VERY INTERESTING!!!!!!

[@tempoprofondo6866]

thank you very much it really helped me to solve this problem ::)

[@fept4043]

I doubt you'll see this but I am sitting the STEP 2 tomorrow, wish me luck!

[@stevemonkey6666]

This was a great journey. I like this video a lot

[@alexmore3865]

I usually don't like youtube math videos (they seem something dumb) but that was beautiful.

[@avishii__________]

Dang this guy is a genius

[@musiconline6849]

Pls make marathon videos on linear algebra

[@ventival]

New to this channel and trying to understending why the pokeball (really good video btw)

[@zacharytoth1065]

It's his microphone.

[@diffusegd]

Ohhhhh I remember this question in one of the random past papers I did

[@nasdpmlima6248]

This was fun. Took me back to 2004

[@TU7OV]

Bruh, I was just working on that exercise from Stewart's Calculus a few days ago.

[@tb-cg6vd]

Some how I passed my STEP 2 exam 30 years ago. I think I was a different me then.

[@nepolionking2393]

He getting power from his ball On his hand😂

[@ryanjagpal9457]

Why is this so hard and complex? Why is it easy for you guys, this looks extremely hard

[@greenland8376]

I was eating frozen pizza while watching this and you now what, it really felt like a six-course menu in a fancy restaurant.

[@blackpenredpen]

😆

[@maxwellsequation4887]

I remember being in the livestream. Good times.

[@suryanshsiddhu007]

My answer for the last question is 8(pie)³/3

[@dxg5659]

Oh, that is awesome. This is why I love math because you can tinker with the equation with methods and properties and find the solution hidden in it. It’s like solving a puzzle. :D

[@MathKhangTran]

I'm from Vietnam. I like your lecture so much 👍👍. But can you explain why in the second part (which you prove the equalities of 2 integrals), you let y = pi - x? How you can think about that idea? I want to understand it clearly. Thanks a lot.

[@superlinux]

Why I feel that you don't raise the bar of challenge, and YOU ARE THE BAR ITSELF ?! Seriously speaking, your channel is a challenge to all computer scientists and mathematicians and physicists.

[@digbycrankshaft7572]

Nice work 👍

[@SabioAC]

If you have a pro version, WolframAlpha probably gave you the same answer. If I enter this in Wolfram Language Integrate[(2x^3-3Pi x^2)/(1+Sin[x])^2,{x,0,Pi}] I get the same answer as you.

[@BalaMurugan-hk2gd]

I am just amazed

[@carlosv.ramirezibanez3305]

Otro gran video, gracias!

[@riadsouissi]

Somehow I miss read the problem and replaced 3 in 3*pi*x^2 with a 2 and the solution didn't simplify nicely 😜 So I ended up with an integral that still got x^2/(1+sin(x))^2 term and tried to solve it. This said, it is in my opinion a more interesting integral to solve. It takes us through integral of log(1+x^2)/(1+x^2) in the interval [0,1] and the solution has the catalan constant in it. To be exact, the integral of x^2/(1+sin(x))^2 is 2/3(pi^2-pi+pi*log(2)+2-4C) where C is the catalan constant.

[@ernestschoenmakers8181]

As an undetermined integral i can only solve it if i change the sinx into a complex notation.

[@shubhrojyotidhara9549]

WELL IT MIGHT BE A STEP PROBLEM BUT IT IS A REGULATION PROBLEM IN JEE ENTRANCE TEST IN INDIA.....ALTHOUGH PRETTY INTERESTING THIS INTEGRAL IS.....I FOUND IT REALLY AWESOME. LOVE FROM INDIA TO BPRP

[@babajani3569]

What is a regulation problem

[@johncalculusmathsclass5998]

Amazing I love this

[@amir78989]

Step questions for me is like Gordon Ramsey on kitchen nightmares sampling the food

[@saharhaimyaccov4977]

Video like this please 👏👑

[@thephysicistcuber175]

I tried it alone. Before watching the video: is it -2/3 pi^3? EDIT: OH MY GUCKING FOD. Total time was 32:45.

[@1timoasif]

You did good! That’s actually about the recommended time you should be spending. There are 6 of these Qs you need to solve and you have 3 hours of time.

[@Akashalomiitbombay]

Am from India Student of mathematics Sir if u are muslim then Assalamu alaikum wa Rahmatullah Love u sir.❤️

[@Akashalomiitbombay]

@@Tiktec then peace be upon you.🤗🤗🤗

[@THNON230]

السلام عليكم اني اتابعك من الوطن العربي...ممكن تضع ترجمة الى اللغة العربية... اعجبني طريقتك بالشرح... استمر👍

[@tgx3529]

Maybe I am wrong, but if I use the substitution x=π/2-2y, then (1+sin(π/2-2y))=2 (cos y)^2, so I have after substitution integral on interval (-π/4;π/4) this integral 1/2 *(-π^3/2)/(cosy)^4. After the substitution z=tg y I have (-π^3/4)* [arctg1-arctg(-1)]= -π^4/8.

[@tristanbiny]

Please teach us to solve simple equation with overcomplicated maths

[@tristanbiny]

yes my goal is to flex

[@ahmadalzyoud4996]

amazing integral

[@benjaminparra4672]

15:04, an edit error 😅, great video!

[@adityaiyer9055]

Question similar to the second question was actually asked in AIEEE(JEE MAIN) once.

[@blackpenredpen]

Oh

[@maximlavrenko1164]

10:40 so you just switched y to x even though y=pi-x??? that doesn't make sense

[@cicik57]

dude you are ready of solving riemann zeta function of 3, do it in some video!!!

[@nicolascamargo8339]

Excelente explicación woow

[@pullingthestrings5233]

BR is a KZbinr ☝🏾PRESENTING to the emergency room.

[@mathevengers1131]

Now do third part(Big boss) without using identities of first and second part.

[@Haalita21]

You do it

[@phoenixninja8872]

Damn that flew right over my head 🥴

[@haider389]

Can you please make a video on "solving jee advanced paper" it will helpful for your channel.. it's a request please.

[@2dlines]

What about the next video on: integral of sqrt({1-sin x}/{1-cos x}) dx (sqrt is the square root function)

[@aashsyed1277]

u are a master of integretion

[@brierhehmeyer1471]

At 19:00 why can you just convert from y to x

[@DesignPredator]

Do some contour integrals

[@youssefaly7067]

wow this is amazing.

[@JG-ne2mx]

No words... just wow

[@mohammadsalkhlayleh3121]

Great Idea ❤❤😍😍😍

[@shivamchouhan5077]

Love your videos Please make videos on jee questions ❤❤❤

[@iceIceCold666]

I LOVE THIS OMG

[@sumanprasad9691]

We in India have to do integrals of comparable difficulties for the JEE Advanced.

[@varunjagan4402]

Yeah cleared it this year......in india just based on face value lots of students can sit for Cambridge and Oxford ....but fees and the interview part of the exam makes us go towards the iits