MIT integration bee qualifier test

[@blackpenredpen]

Hi all, here are some notes/formulas you might find helpful for this video: instagram.com/p/CSxZSlhBPbw/

[@pardeepgarg2640]

2hrs more :(

[@aashsyed1277]

Hiyah!

[@thetruth7105]

Blackpenredpen what kind of math branches did you studied?

[@Caadil6]

Hi dear teacher I want to calculate this Sin4x in terms of sinx The answer will be 8cos³x.sinx -4sinxcosx But I'm confused how to prove that 😕 Please help 🙏

[@julioricardoaguilarsilva1576]

Why log=ln?. Thanks

[@rohanmitra3900]

Great video! Infact my university held its first integration bee, and im glad to say i won! Did not expect it, but i used a lot of tricks that I saw on this channel before!

[@blackpenredpen]

Congrats! 🎊

[@adios04]

which uni do u go to

[@scienceuser4014]

@@adios04 he goes to vermont state

[@chomikhunter]

@@nikhilgond319 what makes you think that

[@SayanGupta4]

​@@nikhilgond319 Stop spreading hate and get a life

[@skylardeslypere9909]

I have an alternate (and faster) solution to Q7. We know cos²x-sin²x = cos(2x) Now we also know that sin(x)cos(x) = sin(2x)/2 So we get the integral of 1/16 * sin(2x)^4 cos(2x) A simple u=sin(2x) will get the job done.

[@enricofuochi2863]

That’s nice but I think you can’t write the answer in terms of 2x as the rules say the answer must be expressed in the original variable

[@skylardeslypere9909]

@@enricofuochi2863 I mean you just get (1/32)(1/5)sin(2x)^5 so if you now plug in sin2x=2sinxcosx again you have an expression in terms of your x

[@mnek742]

@@skylardeslypere9909 @Enrico there is no problem with both versions of the antiderivative, with or without 2x the answers are equally acceptable as they are both in terms of x. If they said u^5/160 + C where u=sin(2x) that would be correct but not acceptable, but there's nothing wrong with 2x occurring in the answer

[@aryanbakshi9876]

Yup same method

[@teslaaf5830]

@DhruvK13 are u from india which university u in?

[@diablo6250]

I really like how you use terms like "U world" and "Complex world" I always find high level math magical, and this really adds to it

[@ガアラ-h3h]

This isn’t high level tho I mean I could solve all them and my friends probably can do and I’m in 10th grade and we’re not even from Asia it’s really basic tbh even the 1/(tan^pi x +1) can easily be done by king rule

[@Marctf1]

@@ガアラ-h3hgood for you and your friend

[@CanadaGoose-ks4eh]

@@ガアラ-h3h You're just better then I guess

[@artchicken1376]

@@ガアラ-h3h ok....good for you. Some people enjoy this kind of stuff.

[@Miftahul_786]

@@ガアラ-h3hshush bro boasting and downplaying other people’s knowledges makes you really dislikable

[@thatkindcoder7510]

I actually learnt quite a bit from this, thx for the video

[@sandglass9928]

For Q11: If you do Integration by parts at the first and then U substitution, that's it. When you do integration by parts it results: -arcsinx/2x² + 1/2 Integral(1/(x²√(1-x²))dx) and the integral can be written as Integral(x^-3[x^-2-1]^-½dx) and with u=x^-2-1 it's done. 😅

[@flix7280]

First one was pretty simple,u substitution is faster than transforming it into an another integral

[@forcelifeforce]

You must mean to write grouping symbols around 2x^2 if it is in the denominator.

[@TinyMaths]

Believe it or not, your 100 integrals in one video helped me figure out a lot more of the MIT Integration Bee (might have been 2005 or 2006) questions that I could have hoped to answer as I followed MIT's video. Your examples helped expose a lot of the holes in my knowledge which prepared me for the Bee (I and I can't wait to dig into this video.

[@mohammadalkousa2856]

Do you have questions of MIT Integration Bee from 2000 t0 2010? If you have, I will appreciate you!

[@mohammadalkousa2856]

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!

[@skylardeslypere9909]

For the first integral, we don't need to worry about x being negative, but that doesn't exclude log(x) from being negative. So the solution should actually be log(2)log(|log(x)|)+log(x).

[@benoist13]

Yes ! In France, if you don't put the absolute value into the ln function, then you don't get full mark !!

[@mnek742]

You're right about the absolute value! Not sure but I think for the MIT Integration Bee the rules indicate that +C and absolute values for logarithms are not required only because it is understood that they belong in the solution.

[@skylardeslypere9909]

@@mnek742 Ah, that could definitely be the case. I'm not from the United States so I have no idea :) I'll take your word for it

[@AviroMusic]

Absolutely

[@jarkola9334]

for question 11 you can just use ' by parts ' directly by taking arcsin(x)/x as first function and 1/x^2 as second and it will simplify beautifully.

[@doom0311]

bruh

[@bastienhumbert255]

Hi! For the 6, we can write: f(x)=sqrt(xsqrt(xsqrt(x...) equivalent to: f(x)=sqrt(x × f(x)) equivalent to: (f(x))^2= xf(x) equivalent to f(x)=x and integrate x

[@hassanniaz7583]

I did Q2 like this: Let, u= e^x + 1 => du=e^x dx => dx= du/(u-1) Integral becomes ∫ du/(u*(u-1)) from 2 to infinity By partial fractions we get, -ln(u) + ln (u-1) from 2 to infinity =ln ((u-1)/u) =ln (1-1/u) By putting limits we get =ln (1-0) - ln (1-0.5) =0-ln (2^-1) =ln(2)

[@4fgaming925]

same

[@aryirfan2427]

Sir, you look like you are gradually progressing into becoming the sensei of mathematics!

[@peterburbery2341]

I really like practicing integration skills! I would like to see another video on another year's integration bee.

[@mokouf3]

First question: Even if those log are of unknown base (but must be same, since they are both written as log), result is still the same. log(a) / log(b) = log_b(a). So we can divide both logs by log(e), making them both becoming ln.

[@migabok.evariste1359]

On question 7 you can also use the indentity sin^4(x)•cos^4(x)= 1/16 sin^4(2x) Then the integral becomes 1/16$ (sin2x)^4 • cos(2x) from here you can make a u-sub. Put u=sin(2x). And finish it off. 😄😆✌👍

[@jonathanhanon9372]

For 11, I did u csc^2(u) cot(u) du IBP: u = u dv = csc^2(u) cot(u) du v = -1/2 csc^2(u) Integrating u dv = -1/2 csc^2(u) * u + 1/2 integral csc^2(u) du = -1/2 (u csc^2(u) - cot u)

[@sherlockjunior8612]

In Q11, what if we directly apply integration by parts, We get: (1/2x²)arcsinx +(1/2) ∫ x⁻²dx/√(1-x²) Here, we have to just calculate the integral: ∫ x⁻²dx/√(1-x²) If we take x² common from the square root in the denominator, we get: ∫ [x⁻²/x√(x⁻²+1)] dx --> ∫ [x⁻³/√(x⁻²+1)] dx Here if we do U substitution of x⁻²+1=t, we get a direct integral of (-1/2)∫dt/√t I think this is a much faster way than first substitution of x=sin(u) and then applyind DI method! Thanks 😇✌🏻

[@adityak1231]

For question 13, using King's rule is better: ∫f(x)dx (a->b) = ∫f(a+b-x)dx (a->b). I = ∫sin(sinx - x)dx I = ∫sin(sinx + x)dx 2I = ∫2sin(sinx)cos(x)dx sinx = t 2I = -2cos(sinx) (0 -> 2π) I = 0 Even for question 15, using King's rule and adding the integrals gives 2I = ∫dx (0->π/2) I = π/4 The tanx part just cancels out.

[@xinpingdonohoe3978]

I did 13 with the king property too. Given the cyclical nature of sin, and the fact that every x was in sin, I didn't even put in the 2πs. I just had the integral of sin(sin(x)-x) = integral of sin(x-sin(x)) and since sin is odd, then I=-I if and only if I=0.

[@Priyanshu-q7s]

the last question can easily be solved by using gama function because if we just substitute x^4 to y and the do some simple algebra involving calc then we will get integration_0 tp inf_0.25{y^(1/2)e^(-y)}dy which is in gama form. 1/4{gama(1 + 1/2)} = 1/4 * 1/2 * gama(1/2) = 1/4 * pi^(1/2)/2 = pi^(1/2)/8

[@holyshit922]

7 Maybe double angle would be better 1/16 Int(sin^4(2x)cos(2x)dx) and simple u substitution u = sin(2x) 11 I would calculate it by parts In first integration by parts I would get rid of arcsinx then I would rewrite integral Int(1/(x^2sqrt(1-x^2))dx) as sum of integrals Int(sqrt(1-x^2)/x^2,dx)+Int(1/sqrt(1-x^2),dx) and integral Int(sqrt(1-x^2)/x^2,dx) again by parts 13. Substitution u = Pi-x and we will get integral of odd function on interval symmetric arount zero 14 If we want to get rid of summation symbol we can use formula fo sum of finite geometric sequence 20. For this rat race this one is quite quick with Gamma function 1/4*Γ(3/2)

[@blackpenredpen]

Nice suggestions!

[@alberteinstein3612]

10:15 if you were to put 1 into that ln argument, you would get ln 0, which is negative infinity. You need to manipulate and then do L’Hôpital’s Rule. Luckily, it would still work out to 0 :)

[@sadececansu9]

Hello, This is where I confused. Could you please give further explanation? The way he solved was wrong? How should we do L'H? Thanks in advance......

[@alberteinstein3612]

@@sadececansu9after he solved the integral, he still had to plug in the upper and lower bounds. I noted that he needed to use log laws and then L’Hôpital’s Rule to actually evaluate these logarithms at x=1, since technically these are the limits as x approaches 1. Hope that helps

[@alberteinstein3612]

@@sadececansu9L’Hôpital’s Rule is used when a limit, via direct substitution, yields either 0/0, inf/inf, or -inf/-inf. It’s when you take the derivative of both the numerator and denominator of the limit and then reevaluate. The limits will approach the same value

[@sadececansu9]

@@alberteinstein3612 Thank you for your respond :) I have another question could you please explain it as well For question 3 how could the teacher find u*lnu because he led u=lnx isn't logx equal to lnx/ln10 ? I said u=logx so lnx=ln10*du then the new interval became 1/ln10 to e/ln10 And for the question 8 how it became ln(x^2+1) but not ln(x^2+1)/ln10 Thank you....

[@alberteinstein3612]

@@sadececansu9 always set u=ln x. The derivative is tens of times easier than log x. Substitute ln for log whenever you can

[@humhum3987]

For 7, we could just develop (cosx+sinx)(cosx-sinx) into cos2x, and sin^4xcos^4x into 1/16*sin^4(2x), then we do u sub : theta = 2x, we have to inetgrate sin^4(theta)cos(theta)/32 which is sin^5(theta)/160 than we develop theta into 2x and we get the answer : sin^5(x)cos^5(x)/5 + C

[@sayantanmazumdar9371]

in the first question u can also do IBP with v=1/x and u=log(2x) /log(x)

[@stratonikisporcia8630]

I don't know what he's doing but I know he's doing it hella good

[@KATTAR_HINDU_AKANDBHARAT]

14:15 here if you multiply divide by 16 it will be integral of sin⁴2xcos2x now put sin2x =u then integrate Way easier

[@jimmykitty]

I'm writing eagerly for the Premier ❤

[@aashsyed1277]

Me too

[@jimmykitty]

@@aashsyed1277 Wow!! How r u? 🥰

[@aashsyed1277]

@@jimmykitty :) fine

[@jimmykitty]

@@aashsyed1277 Glad to hear, kid! 🥰😍❤

[@jimmykitty]

@Corwin Crooks What's this? 🙌

[@justinpark939]

Q11: When we do integration by parts, let u = arcsin and dv = x^-3 Then, we get the UV, which if we evaluate, get 0. The integral part is 1/x^2sqrt(1-x^2) times 1/2 (which I factored out). If we let x=sin(theta), then dx=dthetacos(theta), which the cos term cancels with the radical term (by pythag identity). Therefore, we get 1/sin^2(x) which is csc^2(x). The antiderivative of csc squared is -cot and pluging in the limits yields -1. therefore, the answer overall is a half. Also, if I had known the antiderivative of that integrand to be an arccot, I probably would have been more sure and not substitute out of blind faith.

[@dummydummy5268]

Thank you so much sir, the Di method changed my life

[@randompersonasdf]

for 13) you could also use the substitution u = 2pi - x and you will find I = -I so 2I = 0, I = 0

[@al3diumgd310]

I have a faster solution to Q11: Not going to write it down but essentially integrate by parts, u=arcsin x and dv=x^(-3) it converts to an integral with only powers of x's and it is more handable

[@unkennyvalley287]

wont that give an 1/(x^2 sqrt(1-x^2))? seems troublesome to integrate tbh

[@AquaticDot]

@@unkennyvalley287 It does, but it only takes a simple trig substitution; taking x=sin(u) gives an integral of cos(u)/(sin^2(u)×cos(u)) or the integral of csc^2(u). This is -cot(u) which yields the desired result.

[@jaja47_coolness]

Dang, good to know they see the common log format as the natural log format :/

[@space1607]

This is actually very helpful, I’m a sophomore taking ap calc right now and this video will definitely help me with the class, thanks!

[@blackpenredpen]

😎!!

[@akbaer60]

2:40 It should've been ln(2)ln|ln(x)|+ln(x)+C, because the integral of 1/x dx is ln|x| +c not ln(x)+c

[@nischalpandey201]

Idk why the students (viewers) ain't know about bprp or wtf is the reason why his subscribe is just 750k instead he deserves 2M+......

[@mathevengers1131]

What do you think about my channel?

[@haithambnyslameh2865]

Q7 Use (sin(2x)/2)^4=(sinx cosx)^4= sin^4x cos^4x That's will be easier

[@holyshit922]

I also would calculate in that way

[@balakumarank2006]

The 6th one can be done by substituting w = the given function , then we automatically have w^2 = x.w then x = w then by integrating we have x^2 /2 + c

[@ChaoticNeutral6]

I don't think question 7 was there just to try and slow you down, it looked like a test of trigonometric identities to me. You could have done it faster by using the double angle identities to simplify the expression to (1/16)*sin^4(2t)*cos(2t) and then just use the fact that sin^n(x)cos(x) integrates to sin^(n+1)(x)/(n+1)

[@adandap]

Yes, this!

[@ガアラ-h3h]

Q20) use u = x^4 then see it’s gamma function

[@Train_Sounds]

I think this round microphone ball and him are glued together.

[@TheMichaelmorad]

Q7 can be solved by. noticing that the expression under the integral cam be written as 1/16*(sin(2x))^4cos(2x) and then you have an easy u substitution

[@slavinojunepri7648]

MIT integration bee is an interesting competition. Thanks for these sample exercises as they demonstrate how well candidates must be prepared.

[@mathevengers1131]

3:48 that bring back memories of you with Dr. πm.

[@jVt1306..]

Yeah, you could just split arc sinx and 1/x³ seperately then apply by parts . Would be easy from all these different substitutions . 😊

[@DaveyJonesLocka]

In 4, am I the only one who wondered why he glossed over the fact that (1-x)ln(1-x) is indeterminate as x approaches 1 from the left? I mean, that limit is zero, and hence saying it’s zero just because 1-x is zero is technically correct, but disregards the improperness of the integral.

[@arthurkangdani2414]

Yeah this one is also got on my head because the ln(1-x) one, because if you see the graph, this function is obviously discontinuity at x=1 which makes the logarithms undefined by ln(0) so I still don't understand how it becomes 0. Can anyone share how?

[@fundraiser1907]

Q7 Cn solved way faster here's how you do it multiply and divide by 16 and the integral reduces to (sin2x)^4 cos(2x)dx substitute sin2x=u and you will end up with 1/32 integral u^4 du the resulting ans is same as the one mentioned in the video

[@youngmathematician9154]

Here is a way to solve Q13 analytically (∫(0 to 2π)(sin(sin(x)-x)dx) : Let u=x-π=>x=u+π=>dx=du. Our integral will then become ∫(-π to π)(sin(sin(u+π)-(u+π)))du. By the angle addition identities, sin(u+π)=sin(u)cos(π)+cos(u)sin(π)=-sin(u). Therefore, our integral is equal to ∫(-π to π)(sin(π-u-sin(u)))du= ∫(-π to π)(sin(π-(u+sin(u))))du. By the angle addition identities, sin(π-u)=sin(π)cos(u)-cos(π)sin(u)=sin(u). Our integral is now equal to ∫(-π to π)(sin(u+sin(u)))du. Let f(x)=sin(x+sin(x)). Observe that f(-x)=sin(-x+sin(-x))=sin(-x-sin(x))=-sin(x+sin(x)). This means f(x) is odd. Therefore, the integral we desire will be equal to 0.

[@amitarajesh583]

for ques 6 let y=sqrtx sqrtx sqrt (x) ..... y=sqrt(xy) hence y^2 = xy and considering y will not be constant y = x so integral(x) = x^2/2 + c

[@kamalsaleh6497]

For Q6 you could just set f(x) to sqrt(xf(x)) because of the infinite product. Solving you get x

[@Happy_Abe]

Took an integration bee in 2019 in Brooklyn College and won a round It was much easier and I was just learning Calc 2 then and I managed to win a Pi day shirt We have the bee on pi day lol

[@blackpenredpen]

Nice!!

[@mmpcc4270]

in the 3rd integral you forgot the factor 2 but all the love one of the smartest on the internet

[@justinpark939]

I solved quite a few of these but there is no way I can do 1 per minute. The guys who even do the qualifier are amazing!

[@brendanfay5140]

14:43 you can just use double angle identities and do this way faster amazing videos btw I love this stuff

[@EduardoViruenaSilva]

Integral 13 can be justified by integrating from 0 to pi and then from pi to 2pi. In the second integral a change of variable can be made: u=2pi-x. Now, by using sin(a-b)= sin(a)cos(b) - cos(a)sin(b) the second integral can be reduced to the negative of the first one.

[@ChuiKing]

14:08 question 7 i think it would be easier if we do: 　　sin4 x *cos4 x *(cos2 x -sin2 x) dx 　　= (sinx *cosx)^4 *(2 *cos2 x -1) dx　　　[ sin2 x + cos2 x = 1 >> sin2 x = 1 -cos2 x ], [ cos2 x -sin2 x >> cos2 x -1 +cos2 x >> 2 *cos2 x -1 ] 　　= (sin(2x) /2)^4 *(2 *cos2 x -1) dx　　　[ sin 2x = 2*sinx*cosx >> (sin 2x)/2 = sinx*cosx ] 　　= sin4(2x)/16 *(cos 2x) dx　　　[ 2 *cos2 x = cos 2x +1 >> 2 *cos2 x -1= cos2x ] u = sin 2x du = 1/2 *cos 2x dx 　　sin4(2x)/16 *(cos 2x) dx 　　= 1/16 *u^4 *cos(2x) *2 *1/cos(2x) du 　　= 1/8 *u^4 du 　=> 1/8 *1/5 *u^5 　=> 1/40 *sin5 x　+C //

[@giggitygiggitee1956]

29:55 can be further simplified as (sqrt 8)sin (1+0.25pi) - 2 using sin x + cos x = (sqrt 2) sin (x+0.25 pi)

[@ammardian]

I have a faster solution for question 4, a very unknown formula can make it quicker. arctanh = ln((1+x)/(1-x))/2. This means that you can simply find its integral which is 1/(1-x^2), and hence calculate the solution much quicker.

[@chaitanyakatti463]

The integral of arctanh as you mentioned it wrong, and its actual correct formula is quite hard to remember.

[@ammardian]

​@@chaitanyakatti463 Yep I got it wrong and in the end is quite hard to remember

[@waltermartinez555]

Fun fact: in the integral Q10 the answer could be: (1/2)*ln((2^n)*(x^2)+(2^n)x+(2^(n-1))) with n being a real number, if you don't believe me, try to derive it

[@saivivekpeta1696]

The 13th one will be done by taking x as a+b-x and subsituting, we will directly get zero

[@aashsyed1277]

Sir thanks for inspiring me to make me create a channel......my first video will probably be uploaded on the end of 2021 .......sir thanks a ton!!!!!!!! And once this premiere starts you will be helping me a lot por more than before....MIT integration bee is hard..... But i think you will make it ez........ Thanks for your past videos a ton!!!!!! You deserve 10 million subscribers!!!

[@jimmykitty]

Comment your name on my KZbin channel right now 😁 okay?

[@aashsyed1277]

@@jimmykitty no comments on your channel

[@jimmykitty]

@@aashsyed1277 Comment your name in this video : kzbin.info/www/bejne/Y3OtYo1rn7yapsU

[@aashsyed1277]

@@jimmykitty oh dear......... no people still might see comments...i will tell u one day...

[@jimmykitty]

@@aashsyed1277 Click on this link : kzbin.info/www/bejne/Y3OtYo1rn7yapsU And then drop your name by commenting right now...

[@simran1-08]

Q8, you have to divide by ln10

[@hassanniaz7583]

So I did Q6 in a very unusual way like this: Let, u=sqrt(x*sqrt(x*sqrt(x....))) =>u^2=x * sqrt(x*sqrt(x*sqrt(x....))) =>u^2=x*u =>u=x So, du=dx The integral becomes, ∫ u du =u²/2+c Since u=x =x²/2+c But is my method correct? Comment if anyone finds a mistake.

[@Zephei]

This is also what I did. To fully justify this method you would technically also have to show that u converges for x>0.

[@TheCrashtestCZ]

Amazing video, I really can't stress enough how much I enjoy these!

[@Ahmad-Saad-]

Frist problem could be like that Integratin{(1/x)(log(2x-x)}dx = int{(1/x)(logx)}dx = (1/log |e|)(log|x|) + c

[@TheRageEagle]

I managed to do like 14 of them by myself but definitely not in the time limit hahaha

[@hassanniaz7583]

Same. I did 15. Couldn't do Q9, 11, 13, 14 and 20 on my own.

[@larsb.1972]

@@hassanniaz7583 no one asked

[@hassanniaz7583]

@@larsb.1972 No one told you.

[@4fgaming925]

lol bro salty

[@donaldlogan5310]

As someone who has 0 idea what is going on this is very fascinating and confusing lol

[@derraumdeuter2109]

You're not alone😔

[@unkennyvalley287]

for 13, i realised that using the King property of integration works: int f(x) from x=a to x=b is the same as int f(a+b-x) from x=a to x=b.

[@kingarth0r]

I may try the MIT integration bee someday

[@alejrandom6592]

10:08 shouldn't 0*ln0 be undetermined?

[@ガアラ-h3h]

I’ll explain how to do the 1/(tan^n +1) => int cos^n/(sin^n + cos^n) now use kings rule for the boundaries and you get I = sin^n/cos^n + sin^n add them up 2i = int 1

[@hydra_brothers_gaming7881]

I think in q7 we could've used cos(2x)*sin^4(2x)/16 and then taken u=sin(2x)

[@forgive_me_for_being_unfunny]

The Shirt u are wearing help me really in my exam :)

[@akolangto8225]

The best calculus teacher!!! From the philippines

[@uberless1]

44:50 That 0-0 looks like he's seen some stuff...

[@imabstrong]

For 13 I just did change of var x --> x-π and the fact that sin(x+pi)=-sin(x) to get ∫sin(sinx + x)dx from -π to π. The integrand is odd, so the ans is 0.

[@holyshit922]

These integrals would not be so difficult to calculate without time limit

[@franchello1105]

When i was in school(over 20 years ago), they said ln was base e, log was base 10, and lg was base 2. Is this not true anymore? Also for problem 1, if log is base e, then should you not put your answer back with log instead of ln?

[@rahilkeshav313]

Go watch his video on the confusion of log bases on his second channel called : bprp fast

[@flix7280]

In calculus we mostly use ln with base e until specified

[@franchello1105]

@@flix7280 but my TI-83 uses log as base 10, right?

[@infopek3221]

I believe it differs depending on which course you are in. In CS, people usually use log base 2, therefore log(x) (base not specified) means log base 2. In math, ln(x) refers to log base e, whereas log(x) or lg(x), where the base is not specified, are usually log base 10. They might use log base 2 sometimes, though, it's better to ask.

[@FS-zt6tm]

log can be base anything (log base 10, base 2, base whatever ) and in calculus log base e is what's used. log base e is simply ln

[@plislegalineu3005]

18:50 i'd factor the x² from the denominator

[@nicholasjohnson3542]

Thumbnail question is at 21:17

[@fartoxedm5638]

Q7 - u = cosx - sinx => integral((u^2 - 1)^4/2^4 * u du)

[@Derferder-yt]

Why is this in my recommendations... I haven't even done calculus yet

[@fivestar5855]

When Gaussian Integral appeared you've sparkled with happiness)

[@rahulk5774]

okay i have a question, at 14:33 (which is q no. 7), can we write cos^2(x)-sin^2(x) as cos(2x), multiply and divide by 16, so we would get 1/16 integral(sin^4(2x)cos(2x)), taking sin(2x)=t and then we should be getting the answer as sin^5(2x)/160...is this right?

[@themanofiron785]

2:19 just because x is positive doesn't mean that ln(x) is positive, so ln(ln(x)) is not defined for all positive values of x (in the real world, of course).

[@blackpenredpen]

they didn’t have abs values on the answer key and I was trying to figure out why 😆

[@Someone-cv4bm]

Thanks For Q20 If we use gamma function we will get it in a simple way

[@sagnikdebsarkar169]

What is Gamma function?

[@sagnikdebsarkar169]

I mean he could have just substituted x with sin x or cos x, and of course by changing the limits accordingly

[@154bharshbhanushali9]

Hello I am new to your channel . I liked your videos , I have another method for question 7 , We can use double angle formulaes of cos 2x and sin2x then substitute sin2x = t ; then 2cos2x dx = dt ; and then we can easily solve further ..... By the way, I am from India and I have a fun fact , {In india , our teachers tells us to learn approximately 500 formulaes in Trigonometry only . }

[@erenjaeger309]

500 trignometry formulas??? konsi trigno pdra bhai

[@154bharshbhanushali9]

@@erenjaeger309 Bhai kabhi kundan sir of Physics wallah k lecture dekho Pata chal jaiga Aur bhai joh renowned teachers hain voh questions ko bhi formula banakar yaad karne ko bolte hain Isliye maine toh kiya hai 500 formulaes yaad

[@erenjaeger309]

@@154bharshbhanushali9 ok understood

[@russellkane9439]

Great mr blackpen... many thanks... please MORE bees integrals!!!!! PS... you know what? when you say "product rule" the automatic translator says "PRADA rule" ahahaha

[@mohammadalkousa2856]

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!

[@ashwinrai6783]

Solve JEE Advance`s Integration

[@Ash-pi7ir]

I question 7 you could have made it sin2x power 4 by multiplying and dividing bye 2power4 Then cos²x - sin²x will become cos2x Sin⁴2x.cos2x can be solved easily so you get sin⁵2x/5.2⁵

[@jasonryan2545]

Would it be possible to solve question 5 using partial fractions? I'm aware you've used the method in the video with regards to the time involved, but im still generally curious

[@vardaan0219]

For ques 7 we can actually covert them into higher angle = 1/4 [sin4x cos2x ] and use some trigonometry you write it as 1/8[ sin 6x + sin 2x ]

[@CaoNiMaBi]

Finally, something I can follow.

[@AayushSrivastava0307]

was able to get 18/20 correct in 20 mins, im studying for JEE Adv and this was decent practice , wasnt able to think about Q9 and Q13 in time.

[@fahimnabeel606]

Bro you are good

[@kumarashutosh8937]

in the last question i applied gamma function and got sqrt (pi) / 8

[@thexoxob9448]

The no.1 confused me b/c I interpreted log as base 10. Also why do they make ln notation if people still use log for ln anyway?

[@kavyanshtyagi2563]

i am preparing for jee advanced in india and these questions look very much similar . thnx i did my revision

[@armanavagyan1876]

More MIT integration bee please)

[@theredshadow6696]

For the first question i thought it was log base 10 so i used a calculator and it took me sooo much time😭

[@nocturnalvisionmusic]

What are the chances I revisit this video the day before its third anniversary (8/20/24) - that's absolutely crazy 😧🥳🥰

[@nischalpandey201]

qns no. 17 i have been doing subs thrice and finally got the answer but was quite easy

[@pranjalsingh8017]

Too hard to resist!!