A Cambridge Integral Experience
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- čas přidán 4. 08. 2024
- A calculus integral problem with multiple parts from the Cambridge math admission STEP 2 exam. This video is sponsored by Brilliant: Check them out here brilliant.org/blackpenredpen/
Read more about Cambridge university STEP exams: www.admissionstesting.org/for...
0:00 Cambridge Entrance Exam, STEP 2
0:41 Check out Brilliant to learn more calculus
1:42 part 1, a trig identity
3:48 part 1, integral of 1/(1+sin(x))
5:52 part 2, an integral indentity
12:18 part 2, integral of x/(1+sin(x)) from 0 to pi
15:07 the best part
🔑 If you enjoy my videos, then you can click here to subscribe czcams.com/users/blackpenredpe...
blackpenredpen
“Well why, because it’s on my shirt” no further explanation needed
QED
Source: shirt
proof by shirt
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
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.
Thanks!
@@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.
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
Vincent Abubakar players in Beşiktaş...
Respect++
Best of luck to you.
@@mcalkis5771all the best*
Me at a fancy restaurant
Waiter: Now, to finish your second course, we are you going to serve you... I don't know.
Love this 👍 Classic BlackPenRedPen returns 😀
I would never have thought I would be able to understand every step of such a hard-looking problem
This is awesome. Thank you for this! Please do more of this kind.
Amazing video! I really hope you do some more STEP questions, they’re really fun and it’s very helpful to 2021 step takers!
Great video - step questions are really fun and lots of interesting ones come up. Hope you do some more!
Got this Integral Correct with Perfect Logic!!!
I'm so Grateful....Thanks bprp for the Video❤🙏
Love the STEP content. It's great preparation!
You are doing a great service to all those who love and want to learn math.
this is amazing, thanks for posting!
Bprp : its a cambridge exam question
Me : oh no
Bprp : its *question number 6*
Me : (flashback to the legend) *oh no....*
😄😄
The legendary question number 6 from Australia...
@insert username search on youtube "question number 6 math olympiad"
Oh no no no no no no no
@@alberteinstein3612 even einstein himself afraid from it
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 😂
This was a great journey. I like this video a lot
I also have no idea until I watch your video... You're so brilliant... More power...
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.
The rule you were thinking of is called Kings Rule of Integration… nice solution btw
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.
Wow, I had to prove e was irrational once I entered college, that must be tough for a high school student.
Wow such a complicated integral but beautiful result
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.
Finally step question getting some recognition!!
Tao of our time 👍 great problems solver..you teach me a lot... very unique blackpenredpen
thank you very much it really helped me to solve this problem
::)
Otro gran video, gracias!
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!
I LOVE THIS OMG
THANKS PROFESOR!!!, VERY INTERESTING!!!!!!
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
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
I managed to do it by multiple substitutions!
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
The analogy to the six-course meal at the french restaurant falls in place. How magnificent this integral problem!
Thanks!
Amazing I love this
Nice work 👍
👍🏻Great video
The hardest part of this integral for me was understanding that what I thought is a Greek letter is just how u right cos😂
wow this is amazing.
I got it correct before watching your video! btw this is quite a nice and elegant solution!
👍
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.
Great Idea ❤❤😍😍😍
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
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
@@BCS-IshtiyakAhmadKhan oh I see. Thanks for the insight
Or write sinx/2 as 2tanx/2/(1+ (tanx/2)^2)
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
Sir please make a video of complete set theory with relations and functions. I didn't understand set in maths.
Awesome 👍👌
I am just amazed
It's a hard life in the calculus world!
This was brilliant.
Ohhhhh I remember this question in one of the random past papers I did
Thank you
This is awesommmmmmmmeeeeeee!!!😂
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.
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.
You are amazing
u are a master of integretion
Excuse me,ma cosa serve il calcolo di integrale di x/1+sinx ....che vale π...a cosa serve?thanks
What is the intrigration of square root tan-¹ X.
Pls make marathon videos on linear algebra
amazing integral
Video like this please 👏👑
Very interesting.
I remember being in the livestream. Good times.
No words... just wow
Dang this guy is a genius
New to this channel and trying to understending why the pokeball (really good video btw)
It's his microphone.
Very good
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) ?
What about the next video on: integral of sqrt({1-sin x}/{1-cos x}) dx (sqrt is the square root function)
Excelente explicación woow
Ok ok not bad... but can you do “limit as x approaches infinity of (14x^e)/(log(base x) of 3)?
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.
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.
What is the background track at sponsor message ? Plz tell
Q-6(B) was asked by IIT for Msc. mathematics exam too.
I love you!
BR is a CZcamsr ☝🏾PRESENTING to the emergency room.
I’m confused for part 1 isn’t the integral of 1/(1+sinx) dx equal to tanx - secx + c
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.
He getting power from his ball
On his hand😂
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.
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
I usually don't like youtube math videos (they seem something dumb) but that was beautiful.
I doubt you'll see this but I am sitting the STEP 2 tomorrow, wish me luck!
I was eating frozen pizza while watching this and you now what, it really felt like a six-course menu in a fancy restaurant.
😆
how do you get sec^4u is and even function?
Grande!!!!!from italy
dude great video where to get your shirt ?
Damn that flew right over my head 🥴
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?
how well would residue integration work?
At 19:00 why can you just convert from y to x
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🙇
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.
EPIC!!!!
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.
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.
@@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.
10:34 sir how did you write dy = dx and interchange the upper and lower limits simultaneously didn’t we have dx = -dy…
This was fun. Took me back to 2004
Can you factor 1+(a^3)+(a^3)(x^3) please
Beautiful
Bruh, I was just working on that exercise from Stewart's Calculus a few days ago.
Love your videos
Please make videos on jee questions ❤❤❤
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.
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 :)
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.
@@violintegral had the exact same experience haha
dude you are ready of solving riemann zeta function of 3, do it in some video!!!