“Well why, because it’s on my shirt” no further explanation needed
@yusefzaman52422 жыл бұрын
QED
@MrUtah1 Жыл бұрын
Source: shirt
@m1n3c4rt8 ай бұрын
proof by shirt
@toby.barnett3 жыл бұрын
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
@abubakar-serdiq3 жыл бұрын
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
@remzidag87353 жыл бұрын
Vincent Abubakar players in Beşiktaş...
@mujtabanadeem39012 жыл бұрын
Respect++
@mcalkis5771 Жыл бұрын
Best of luck to you.
@abhirupkundu27783 ай бұрын
@@mcalkis5771all the best*
@finnhogan55253 жыл бұрын
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.
@blackpenredpen3 жыл бұрын
Thanks!
@anandk92203 жыл бұрын
@@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.
@cycklist3 жыл бұрын
Love this 👍 Classic BlackPenRedPen returns 😀
@anshumanagrawal3463 жыл бұрын
I would never have thought I would be able to understand every step of such a hard-looking problem
@krishnanavati37153 жыл бұрын
Great video - step questions are really fun and lots of interesting ones come up. Hope you do some more!
@echohasbinokiller43 жыл бұрын
Me at a fancy restaurant Waiter: Now, to finish your second course, we are you going to serve you... I don't know.
@gigachad24193 жыл бұрын
Got this Integral Correct with Perfect Logic!!! I'm so Grateful....Thanks bprp for the Video❤🙏
@babajani35693 жыл бұрын
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.
@Jasomniac2 жыл бұрын
Wow, I had to prove e was irrational once I entered college, that must be tough for a high school student.
@DevTech03Ай бұрын
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
@ismailberbache87803 жыл бұрын
Amazing video! I really hope you do some more STEP questions, they’re really fun and it’s very helpful to 2021 step takers!
@insouciantFox3 жыл бұрын
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 Жыл бұрын
The rule you were thinking of is called Kings Rule of Integration… nice solution btw
@tfg6012 ай бұрын
@@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.
@parthgupta533 Жыл бұрын
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 😂
@mohammadfahrurrozy80823 жыл бұрын
Bprp : its a cambridge exam question Me : oh no Bprp : its *question number 6* Me : (flashback to the legend) *oh no....*
@mahatmaniggandhi28983 жыл бұрын
😄😄
@fivestar58553 жыл бұрын
The legendary question number 6 from Australia...
@mohammadfahrurrozy80823 жыл бұрын
@insert username search on youtube "question number 6 math olympiad"
@alberteinstein36123 жыл бұрын
Oh no no no no no no no
@mohammadfahrurrozy80823 жыл бұрын
@@alberteinstein3612 even einstein himself afraid from it
@tommc14253 жыл бұрын
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
@jewel84822 жыл бұрын
I managed to do it by multiple substitutions!
@NoName-kj2vf Жыл бұрын
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
@alwayslistening4444 Жыл бұрын
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.
@moinmalik1320 Жыл бұрын
You are doing a great service to all those who love and want to learn math.
@slavinojunepri7648 Жыл бұрын
The analogy to the six-course meal at the french restaurant falls in place. How magnificent this integral problem!
@blackpenredpen Жыл бұрын
Thanks!
@MathElite3 жыл бұрын
Wow such a complicated integral but beautiful result
@seshnarayan79723 жыл бұрын
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-IshtiyakAhmadKhan3 жыл бұрын
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
@seshnarayan79723 жыл бұрын
@@BCS-IshtiyakAhmadKhan oh I see. Thanks for the insight
@varunjagan44022 жыл бұрын
Or write sinx/2 as 2tanx/2/(1+ (tanx/2)^2)
@tkc-g8h3 жыл бұрын
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!
@ekisvioleolivaradamos67013 жыл бұрын
I also have no idea until I watch your video... You're so brilliant... More power...
@yamikira65122 ай бұрын
I saw the thumbnail, started doing the integral, finally managed to evaluate it, then opened the video and realized... it's a step question.
@dirt35543 жыл бұрын
Love the STEP content. It's great preparation!
@Eric-dt7bt3 жыл бұрын
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
@mohammedhaidar32002 жыл бұрын
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 Жыл бұрын
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.
@joshhutchinson70313 жыл бұрын
Finally step question getting some recognition!!
@rob8763 жыл бұрын
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.
@yoyoezzijr2 жыл бұрын
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
@nolimitderrick48223 жыл бұрын
The hardest part of this integral for me was understanding that what I thought is a Greek letter is just how u right cos😂
@dmddjack3 жыл бұрын
I got it correct before watching your video! btw this is quite a nice and elegant solution!
@blackpenredpen3 жыл бұрын
👍
@morischacter10762 жыл бұрын
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.
@colfrancis97253 жыл бұрын
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_wooster3 жыл бұрын
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.
@colfrancis97253 жыл бұрын
@@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.
@Xerkun2 жыл бұрын
This is awesome. Thank you for this! Please do more of this kind.
@reu.mathematicsacademy85662 жыл бұрын
Tao of our time 👍 great problems solver..you teach me a lot... very unique blackpenredpen
@bishalbhattacharjee59303 жыл бұрын
Sir please make a video of complete set theory with relations and functions. I didn't understand set in maths.
@chessandmathguy3 жыл бұрын
this is amazing, thanks for posting!
@goodra49993 жыл бұрын
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 Жыл бұрын
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.
@benoist132 жыл бұрын
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.
@MathKhangTran2 жыл бұрын
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.
@MartinPerez-oz1nk Жыл бұрын
THANKS PROFESOR!!!, VERY INTERESTING!!!!!!
@AdrienLegendre2 жыл бұрын
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.
@nickallbritton37962 жыл бұрын
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.
@narrotibi3 жыл бұрын
It's a hard life in the calculus world! This was brilliant.
@erenozturk5863 жыл бұрын
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 :)
@violintegral3 жыл бұрын
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.
@erenozturk5863 жыл бұрын
@@violintegral had the exact same experience haha
@ginopaperino2608Ай бұрын
this man is powerful
@aalsii3 жыл бұрын
Q-6(B) was asked by IIT for Msc. mathematics exam too.
@stevemonkey66663 жыл бұрын
This was a great journey. I like this video a lot
@fept40432 жыл бұрын
I doubt you'll see this but I am sitting the STEP 2 tomorrow, wish me luck!
@musiconline68493 жыл бұрын
Pls make marathon videos on linear algebra
@tgx35293 жыл бұрын
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.
@diffusegd3 жыл бұрын
Ohhhhh I remember this question in one of the random past papers I did
@tempoprofondo68663 жыл бұрын
thank you very much it really helped me to solve this problem ::)
@maximlavrenko11642 жыл бұрын
10:40 so you just switched y to x even though y=pi-x??? that doesn't make sense
@superlinux2 жыл бұрын
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.
@avishii__________3 жыл бұрын
Dang this guy is a genius
@god-zilla42332 жыл бұрын
I don't really understand dummie variables. Is it because you can replace it with pi without any effect on the definite integral? Because it changes the primitive (if it wasnt definite) no?
@ventival3 жыл бұрын
New to this channel and trying to understending why the pokeball (really good video btw)
@zacharytoth10653 жыл бұрын
It's his microphone.
@ryanjagpal94573 жыл бұрын
That Sin y one didn’t really make any sense cause like you removed the y so why is the y still there?
@sandhuekam14 Жыл бұрын
10:34 sir how did you write dy = dx and interchange the upper and lower limits simultaneously didn’t we have dx = -dy…
@bishanmadanayake76872 жыл бұрын
Please, I don't understand how you considered (pi-x) as “x” straight away in 10:47
@alexmore38652 жыл бұрын
I usually don't like youtube math videos (they seem something dumb) but that was beautiful.
@nepolionking23933 жыл бұрын
He getting power from his ball On his hand😂
@digbycrankshaft75723 жыл бұрын
Nice work 👍
@brierhehmeyer1471 Жыл бұрын
At 19:00 why can you just convert from y to x
@SabioAC2 жыл бұрын
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.
@THNON2303 жыл бұрын
السلام عليكم اني اتابعك من الوطن العربي...ممكن تضع ترجمة الى اللغة العربية... اعجبني طريقتك بالشرح... استمر👍
@maxwellsequation48873 жыл бұрын
I remember being in the livestream. Good times.
@thephysicistcuber1753 жыл бұрын
I tried it alone. Before watching the video: is it -2/3 pi^3? EDIT: OH MY GUCKING FOD. Total time was 32:45.
@1timoasif3 жыл бұрын
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.
@ryanjagpal94573 жыл бұрын
Why is this so hard and complex? Why is it easy for you guys, this looks extremely hard
@giuseppemalaguti4353 жыл бұрын
Excuse me,ma cosa serve il calcolo di integrale di x/1+sinx ....che vale π...a cosa serve?thanks
@nasdpmlima62482 жыл бұрын
This was fun. Took me back to 2004
@TU7OV5 ай бұрын
Bruh, I was just working on that exercise from Stewart's Calculus a few days ago.
@MohitGupta-wg4vt Жыл бұрын
Why dont you use property no. 4, in india it is also called king property
@sumanprasad96913 жыл бұрын
We in India have to do integrals of comparable difficulties for the JEE Advanced.
@varunjagan44022 жыл бұрын
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
@dac8939 Жыл бұрын
My step exams for cambridge 2013 were not too hard. Not much more than A level further maths. Got S, S in both papers with no extra prep.
@2dlines3 жыл бұрын
What about the next video on: integral of sqrt({1-sin x}/{1-cos x}) dx (sqrt is the square root function)
@dantesk18362 жыл бұрын
Didn't understand why is it possible to change y for x in the second integral when he's doing the distribution in step 2
@superhitsuperheroes3 жыл бұрын
What is the intrigration of square root tan-¹ X.
@shivaygupta89013 жыл бұрын
Sir plz add some Împortant payment method for brilliant .. ex . Paytm, upi , plz sir 🙏🙏🙏🙏🙏🙏🙏🙏🙏
@nicolascamargo8339 Жыл бұрын
Excelente explicación woow
@haider3893 жыл бұрын
Can you please make a video on "solving jee advanced paper" it will helpful for your channel.. it's a request please.
@leopoontw2 жыл бұрын
wait hi bprp, how do we come up with the pi - x by ourselves (let's say they didn't tell us to make that sub) ?
@carlosv.ramirezibanez33053 жыл бұрын
Otro gran video, gracias!
@bakashisenseiAnimeIsLove3 жыл бұрын
can you plss solve the integral root(x^3 + 1) dx using a simple method plss...i am stuck on that qns from several days
@violintegral3 жыл бұрын
Non-elementary
@sunandinighosh60373 жыл бұрын
I think it's not integrable...you should try inequalities
@tb-cg6vd Жыл бұрын
Some how I passed my STEP 2 exam 30 years ago. I think I was a different me then.
@jayjayf96993 жыл бұрын
I dont understand why pi - x is used as change of variables, i dont understand the explanation of symmetry, i know what odd and even functions are so i dont understand when u say its symmetry
@Akashalomiitbombay3 жыл бұрын
Am from India Student of mathematics Sir if u are muslim then Assalamu alaikum wa Rahmatullah Love u sir.❤️
@Akashalomiitbombay3 жыл бұрын
@@Tiktec then peace be upon you.🤗🤗🤗
@suryanshsiddhu007 Жыл бұрын
My answer for the last question is 8(pie)³/3
@greenland83763 жыл бұрын
I was eating frozen pizza while watching this and you now what, it really felt like a six-course menu in a fancy restaurant.
@blackpenredpen3 жыл бұрын
😆
@giuseppemalaguti4353 жыл бұрын
Ho terminato....ma questi integrali li sogni si notte?non credo,se no c'è da preoccuparsi
@riadsouissi3 жыл бұрын
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.
@pullingthestrings52333 жыл бұрын
BR is a KZbinr ☝🏾PRESENTING to the emergency room.
@saharhaimyaccov49772 жыл бұрын
Video like this please 👏👑
@shubhrojyotidhara95493 жыл бұрын
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
@babajani35693 жыл бұрын
What is a regulation problem
@darshandesai72253 жыл бұрын
I’m confused for part 1 isn’t the integral of 1/(1+sinx) dx equal to tanx - secx + c
@ernestschoenmakers8181 Жыл бұрын
As an undetermined integral i can only solve it if i change the sinx into a complex notation.
@DesignPredator3 жыл бұрын
Do some contour integrals
@tylerwoodward97253 жыл бұрын
Ok ok not bad... but can you do “limit as x approaches infinity of (14x^e)/(log(base x) of 3)?
@tristanbiny3 жыл бұрын
Please teach us to solve simple equation with overcomplicated maths