from the MIT integration bee! 🐝

[@TechyMage]

Using kings rule u will get I= same upper and lower limit √2cos x/(9+16cos 2x) And after some painful simplification and substitutions u will get the final ans

[@lavneetjanagal]

It can be done much faster if you notice that sin(x)+cos(x) is actually the derivative of sin(x)-cos(x) . In the denominator 9+16 sin(2x) = 25-16(sin(x)-cos(x))^2 .And you have the final form of the integral -1 to 0 1/(25-16 u^2) . you can use partial fraction or trig substitution or contour to do this integral.

[@goodplacetostop2973]

14:45 No homework today but I wish all the best to students about to start their finals.

[@johnbailey8103]

Just finished all of mine, but thank you very much, I love all your post video homeworks ❤️

[@calcul8er205]

u=sinx-cosx allows you to get rid of the numerator then the denominator can be expressed as 25-16u^2

[@megauser8512]

Yep, I saw that method/trick from this page: www.teachoo.com/4856/728/Misc-30---Definite-integral-0----pi-4-sin-x---cos-x/category/Miscellaneous/

[@megauser8512]

And then here's how to get the formula used after that method: 1drv.ms/b/s!AvQhCl0SMET3ygVtkhZcUuC6HhOO?e=fv7ToI.

[@riadsouissi]

I used weierstrass substitution t=tan(x/2), ended up with an intimidating integral of 2(1+2t-t^2)/(9t^4-64t^3+18t^2+64t+9) between 0 and sqrt(2)-1 However, denominator can be factored into 2 second degrees polynomials. After painful manipulations (separating two denominator polynomials in separate fractions), I end up with integral of 1/40((18t+8)/(9t^2+8t+1)-(2t-8)/(t^2-8t+9)) = 1/40(log(9t^2+8t+1/t^2-8t+9). And this gives same result. Painful method but no tricks except the initial substitution.

[@polyhistorphilomath]

This is the way.

[@adityapandey4388]

Lol i got this question in my mid term exam in my school in india.

[@hargunbirsingh6459]

Take (sinx-cosx)= t, differentiate and put numerator=dt the then put sin2x= 1-t^2

[@djvalentedochp]

I love integrals, so good seeing all those transformations till getting to a well known shape

[@ii4511]

You can complete the square for the denominator by +-7, and subbing in the basic identity for cos and sin, then use harmonic from, and finally a substitution

[@mastermt5904]

Also you can try using the identity integrate f(x) from upper limit a to lower limit b equals to f(a+b-x) from upper limit a to lower limit b

[@tgx3529]

One more the solution for this problem is: We can take integral from (sqrt(2)*sin(π/4+x))/(9+16sin2x), then we take the substitution π/4+x=y, we have the integral from (sqrt(2) sin y)/(25-32*(cos^2)y) from π/4 till π/2, after the substitution cos y= k we have integral from sqrt(2)/(5+sqrt(32)k)*(5-sqrt(32)k) where k go from 0 till (sqrt2)/2.....finally we get the same result.

[@riadsouissi]

this is the more natural / less tricky way of solving it.

[@tomasstride9590]

I did not press my approach on to the bitter end which I did before even looking at the video. But I think you can also use this: 9+16sin2x = 9+32sinxcosx= 9-(4sinx-4cosx)**2+16= 25-(4sinx-4cosx)**2 Now use difference of two squares formula and the partial fraction. You will see the part in the derivative of the brackets is more or less the numerator so the integrals are just logs.I am satisfied this works but did not go on to the bitter end.

[@Eldar1on93]

From here it's u=sinx-cosx and (sinx+cosx)dx = du, u := -1...0, so integrate -1...0 (1/25-16u^2) du Great trick, I almost got there but missed (tried with (sinx+cosx)^2)

[@qillerdaemon9331]

Fascinating to watch these videos. Penn starts from an equation with trig functions, and ends up with a solution with a natural log. I'm sure somewhere there is something that connects trig functions with natural log functions, but as a former chemist and now in IT infrastructure, I have absolutely no idea how. I mean, I barely got thru the mess that is physical chemistry. :(

[@BenjersVeggie]

There is! The antiderivative of tan(x) is -ln |cos(x)|, and the antiderivative of sec(x) is ln |sec(x)+tan(x)|, both treated as real-valued functions.

[@polyhistorphilomath]

Or the logarithmic derivative is involved due to the form. It’s a sum of rational functions. Also -i log(cos z + i sin z) = z.

[@paulmuaddib3761]

one way might be this: natural log is obviously related to e and e and sin and cos are related to each other by e^ix = cosx + isinx

[@ramanakv3272]

Sin2x=1-(sinx-cosx)^2 and proceed by substitution method

[@aliasgharheidaritabar9128]

It was so great.thank u for ur wonderful videos

[@berzerksharma]

this is a extended method , you can subtract and add 1 to the denominator, and use 1-sin2x in the denominator as (sinx-cosx)^2 and substitute the sinx-cosx = t and boom you are into the final steps

[@manucitomx]

That was really cool. I hope it comes up in my tutoring next semester.

[@mathssolverpoint6059]

Use t = sinx +cosx

[@tgx3529]

time 8:30. We can take the substitution t=sqrt(25-u)

[@chessematics]

Solve x"+ax=b×cos(wt+j) x=x(t), a,b,w, j are constants

[@Tiqerboy]

Wow. I guess there's no faster way to solve that. Never would have gotten that since my calculus years are too far into the past but it was fun to follow your work.

[@leadnitrate2194]

i like Michael's method, but as others have pointed out, there's a much faster way to solve it (sinx - cos x)² = 1 - sin2x sin2x = 1-(sinx-cosx)² Substitute this as the value of sin 2x and then substitute u = sin x - cosx du = (sin x + cos x)dx which lets you get rid of the denominator leaving you with a very standard integral

[@mohammadalkousa2856]

Great! Recently it was published a book about MIT integration bee, under the title " MIT Integration Bee, Solutions of Qualifying Tests from 2010 to 2023" You can simply find it!

[@whyyat3470]

How was he able to switch the signs of the factors for the partial fraction decomposition at 11:45? The factors are 5-y and 5 + y...

[@William_35]

Didn't he multiply the entire equation by -1 or am I missing something?

[@nournote]

For partial fractions, it would be easier if you multiply only by (y-5) and replace y=5, that would give A directly, and then multiply by (y+5), replace by y=-5 and get B direclty without doing a system of 2 equations.

[@aaaaansh4907]

This question is taught to us for jee preparation as a particular type... I it's awesome

[@nuranichandra2177]

Superbly elegant

[@higgy4596]

Can someone explain the simplification of 7:24? I don’t see how the 1/8 is being pulled out. How can he pull out it from inside the square root?

[@zornslemmon2463]

He pulls the 16 out, which becomes a 4 outside the square root. Now you have (1/32) * (1/(1/4)) = (1/32) * 4 = 1/8

[@higgy4596]

@Zorn’s Lemmon I think I get it. Since he is taking a (1/16), the new expression would be sqrt(1/4)*sqrt(25-u). Then the 1/4 would flip because you would divide 1/(1/4) making the four on top which gives the 1/8 which you can pull out. Thank you very much!

[@higgy4596]

@@zornslemmon2463 Thanks again!

[@talkravarusic8489]

Really fun content ! :)

[@jonathanengwall2777]

At3:22, you have 1/ududx it is ugly but the numerator is gone. Why not solve it?

[@Meverynoob]

After subbing in u he's left with x terms, so he has to convert it into a u term, but the resulting integral has a complicated root so he performs another substitution t to kind of swap over, but that is still complicated so he substitutes y such that multiplying dt/dy removes the root from the denominator and the solution from there is straightforward.

[@jonathanengwall2777]

@@MeverynoobI watched the entire video also, thank you. It appears to be a polar coordinates problem. Canceling from within the u term ruins the u term, I see, but I am commenting on the overworking of a 90 degree rotation. There is no need for a y term, only theta.

[@muzankibutsuji5572]

U can also do this by Sinx+cosxdx/9+16sin2x -(sinx+cosx)dx/16(1-sin2x)-25 -(sinx+cosx)dx/16(sinx-cosx)^2-25 Sinx-cosx=t -dt/16t^2-25 1/4*1/10ln(4t-a/4t+a)(I remember this formula of dx/x^2-a^2 proof is by partial fraction) Thus put proper limits from -1 to 0 u will get the answer

[@winky32174]

Another good one.

[@monikaherath7505]

At what point does integration just become applying a bunch of known techniques and equation manipulation without really feeling or understanding what the integral means? I understood every step in the video but I had to go and literally plot out some of the graphs of the functions shown to get any hint of intuition to what it was trying to do.

[@himanshuverma1580]

First apply king property, then put sinx=t and then it become a simple formula based question

[@himanshurathore2843]

Yes this will be solve in short method.

[@arihantbhansali]

Bro your solution is best

[@abderrahmanelhamidy1620]

nice solution

[@arpitdas4263]

I did this in a different way.I used the f(x)=f(a+b-x) property to simplify the integral and then expand sin and cos to obtain sin/1.41 (value of root 2) in the numerator.Then I substituted cos2x=(2cosx)^2 - 1 and finally substituted cosx as u to get an integral in the arctan form When I entered the value of this arctan in Wolfram Alpha,I obtained the same numerical expression being obtained through the logarithmic form,so I think my method works as well

[@dingo_dude]

what a delicious integral

[@adityapandey4388]

I got this question in my midterm in class 11 lol

[@muzankibutsuji5572]

U could also have done by putting sinx +cosx=t sin2x would become t^2-1 then put t^2 =y

[@rbdgr8370]

But then dx=dt/√2-t^2 would have posed problems

[@muzankibutsuji5572]

@@rbdgr8370 Sinx+cosxdx/9+16sin2x -(sinx+cosx)dx/16(1-sin2x)-25 -(sinx+cosx)dx/16(sinx-cosx)^2-25 Sinx-cosx=t -dt/16t^2-25 1/4*1/10ln(4t-a/4t+a)(I remember this formula of dx/x^2-a^2 proof is by partial fraction) Thus put proper limits from -1 to 0 u will get the answer

[@muzankibutsuji5572]

@@rbdgr8370 ignore the above msg

[@zhusan2dui]

That is how I did this integral.

[@oder4876]

Man you have an factorisation error (25-y^2) = (5-y)(5+y) 11:40 not the opposite lol😉 but he will gave the same result at end A=B permutate but i will give you like 👍 what ever, to your hard word keep going and good luck

[@faus2417]

he canceled that effect by multiplying the other side by -1

[@chessematics]

Do Weierstaß substitution

[@mastermt5904]

I think you can use the t-formula for this question, that would be easier

[@williamwarren5234]

Is this substituting t=tan(x/2)?

[@mastermt5904]

@@williamwarren5234 yes, you can also find sinx and cosx in terms of t

[@paulconnor1040]

Tried it that way, it's not easier. Bigger, longer and more fiddly, lots of manipulation of surds of the form a+bsqrt{7}.

[@marcososorio1885]

Steven dice que no es trivial

[@factorization4845]

If you start with u=pi/4-x, then everything just simplifies quickly

[@infinitygaming5808]

I thought about it

[@geometrydashmega238]

To be honest this is the first substitution I did but then I got stuck. It wasn't easy for me :/ Edit: I got it. Just had to substitute cosu = t and do a bunch of calculations

[@spartacus8875]

Thank you....

[@anthonyguerrera191]

Is it just me or is the PFD flipped

[@VerSalieri]

This was super fun!! Thank you.

[@MuslimStoic123]

It's is a easy problem I solved it in only one substitution And using king property Good content

[@Yk.T]

Why MIT int calc competiton is called Integration "Bee" ?

[@TJStellmach]

It's an English term used for a competition (especially, but not necessarily, a spelling contest) or a gathering to pool labor (e.g. a "sewing bee").

[@Pauldyke]

14:14 ln(5) - ln(-5) is not zero.

[@theloganator13]

Absolute value

[@Pauldyke]

@@theloganator13 Yh I noticed after I posted.

[@2kworld9]

Just put sin x- cos x=k its a 2 minute question

[@fancpleq6031]

Cosx-sinx should be substituted as from 0-π/4 cosx is greater than sinx

[@ardak4000]

Wow I couldn't do it if sin2x=2sinx.cosx wouldn't given

[@giuseppemalaguti435]

1/40ln9....io lho fatto ponendo al posto di x......(pi/4)-x......

[@legendarydisciplegaming4316]

Need more problems on ramanujams

[@user-A168]

Good

[@abraarmasud9194]

Had this in our finals last year. Solved using king's rule. It'll be way shorter!

[@stapler942]

Honestly I think it depends on the exam because if you happen to know King's Rule and they want you to show all the steps, you might have to derive King's Rule first.

[@abraarmasud9194]

@@stapler942 our teacher was okay with it. So, win!

[@pandas896]

Very easy

[@biolinux2307]

Matemáticas 😎👍

[@saiteja3885]

May I know why mit ask too easy problems mainly in mathematics?

[@divyanshaggarwal6243]

in the integration bee, the questions are usually easy but the point of the competition is to check the speed of the contestants.

[@polyhistorphilomath]

@@divyanshaggarwal6243 which seems pointless. I think anyone in the bee or watching this video (willingly) would be capable of writing a program to express such solutions in terms of hypergeometric functions. The machine execution will be faster than manual calculations. Why bother?

[@earlyaspirant3301]

It would be way easier if: 1.we write sin2x=1-(sq(sinx-cosx)) 2.Put sinx-cosx=t We will have dt=( sinx+cosx)dx same as numerator.

[@debojitdeori4837]

Classic question from indian ncert for 12 standard students

[@advaykumar9726]

Yes

[@anjaneyasharma322]

I have said so many times Trigo calculus is nonsense in its present form. Both lists of integrals and derivatives are useless. Take integrals of sinx from 0 to pi and integral of cosx from 0 to pi. If you draw the graphs you will find the integral should be equal as taking integral from 0 to pi/2 the result should be 2. But not so because the reaches are not aware of this mistake. Unknowingly fooling is going on from Newton Leibnitz times to this day. This is happening at MIT. So you can imagine the picture in rest of the world

[@TechyMage]

Dekha apne laparwahi ka nateja

[@aaryunik]

Hello!

[@legendarydisciplegaming4316]

Hi

[@gamert2525]

Too easy problem

[@CM63_France]

Hi, For fun: 1 "so the first thing that we would like to do is", 1 "so let's go ahead and do that", 3 "let's may be go ahead and", including 1 "so let's may be go ahead and use that", and 1 "so let's may be go ahead and write that", 1 "now I'll may be go ahead and", 1 "I can go ahead and", 1 "we need to go ahead and", 2 "ok, great", 1 "great",

[@blackdeutrium746]

❤️

[@shreyashrao9583]

I did it in 5 mins 👍 iam a 12 th grade student

[@maxwellsequation4887]

Third?

[@liviuliviu58487]

i see 1 dislike. why would someone dislike this video?

[@237_alifamirudin2]

ln 3 = 1

[@earlyaspirant3301]

@@angelmendez-rivera351lol you didnt the joke baby.

[@tobycooper4639]

Bruh as blackpenredpen would say, you're still in the y world. Gotta get back to x

[@ShamJam85]

to much maths! tooo much tedious isnt it?

[@darreljones8645]

The answer, rounded to two decimal places, is 21.97.

[@ucduong2102]

Nice bro but i have a better way to solve

[@prabhatsharma5751]

2nd❤

[@shd04109]

I have only one suggestion. SPEAK UP PLS.

[@Te4mRyouko]

Yeah, the volume in his videos is kinda low compared to other videos on KZbin.