Long story short. S1- apply king property. Add 1 and 2. S2 -separate log(a/b) to loga -log b S3 - get sin2x term in terms of I by substituting 2x =t voila😎😎

@gytricky4987

6 ай бұрын

Ashish sir student?

If u are thorough with formulae and problem solving techniques, Solution starts at 8:45

@flaviasaldanha1537

3 жыл бұрын

Thank you

@kshitiz6376

3 жыл бұрын

Thanks man. I would've rage quit this video had I not seen your comment.

@KA-ff4uq

2 жыл бұрын

Tysm

For the record, f(x) = log[sin(x)] has no elementary anti-derivative.

@brocks7191

4 жыл бұрын

I tried doing the sine infinite product proof, it wasn’t pretty. What would be the closest elementary integral?

@gagannayyar990

4 жыл бұрын

@@brocks7191 I saw a very creative approach on math stackexchange using complex analysis. I have only studied complex analysis in high school hence didn't understand the method but it was intriguing.

@me_hanics

4 жыл бұрын

Nevermind, i think my comment has a flaw

@stewartzayat7526

4 жыл бұрын

I wonder how people can prove that there is no elementary anti-derivative

@nachiketsharma4507

3 жыл бұрын

Yeah, but computers have come up with special functions in which we can compute the value of any defined interval

i tried the sine infinite product formula and yeah don't do that.

@nachiketsharma4507

3 жыл бұрын

Dude, that is stupid

@kishorekumarsathishkumar1562

3 жыл бұрын

@@nachiketsharma4507 it's not that dumb, you can use log properties and take out the summation outside the Integral, which is normally very useful...

I got this question a week or two ago for homework and it was relatively easy (they gave us a clue to use the double angle fomula)

haha i gotta appreciate how you slaughter the pronounciation of integral :p

you can also get the sin,cos being the same integral from symmetry as well, without the u sub.

This integral is evaluated by advanced methods in Lars V Ahlfors's book Complex Analysis (Third Edition, 1979, pages 160-161). I suppose the author overlooked the elementary method!

@user-uh9bo2im1h

10 ай бұрын

Well if it’s without bounds you have to do it with complex analysis because this works due to symmetry

I have tried to do it by myself and after many tries with substitutions, Feynman's technique, I went exact same way as you did, but than I got stuck when I go to integral of ln(sin(2x),fine we have it, but how to come back to original integral, I gave up and watched the video and I feel so stupid by not noticing simple u=2x substitution. Thank you for interesting puzzle anyway.

@LetsSolveMathProblems

5 жыл бұрын

Ahh, you were so close! =)

Man u r waking up the spirit of maths in me

Thanks sir, your explanation was clear, I think that it doesn't exist another way to calculat this integral except for your idea, so ... thank you very much.

@user-uh9bo2im1h

10 ай бұрын

It is possible with complex analysis and kings rule

Awesome exercice very elegent .

Very good video ! Thanks :)

Try it with Pn=product from k=1 to n-1 of sin(kpi/2n) (Starting with Pn^2)

Soooo what do we do if we didn’t know how the graph looks like?

@nikolaivii5766

5 жыл бұрын

@Blan Morrison I tried this on Wolfram Alpha with a few other curves, and it seems that natural logging a function preserves the function's symmetry as you state. Why is this? It doesn't sound like it need be the case, and yet it is?

@AymanSussy

5 жыл бұрын

integral from 0 to Pi of ln(sin(x)) is I ( our integral ) + the integral from Pi/2 to Pi now u should prove that the second integral is equal to I as well ( use T=x - Pi/2 )

@93683409

4 жыл бұрын

@@nikolaivii5766 doesn't matter whether it's ln(X) or not. Say f(g(X)), as long as g(X) is symmetrical, f(g(X) will also be symmetrical. Because you can't get two different values of f when g(X) is the same.

@nachiketsharma4507

3 жыл бұрын

Looking at the graph doesn't make a difference

@davidbrisbane7206

3 жыл бұрын

Show that: Integral ln(sinx) from x = 0 to ℿ/2 = integral ln(sinx) from x = ℿ/2 to ℿ. Now sin(ℿ-x) = sinx, as sin(ℿ-x) = sinℿ*cos(-x) + cosℿ*sin(-x) So, sin(ℿ-x) = 0*cos(-x) + (-1)*sin(-x) = 0 + (-1)*(-sinx) = sinx. Consider the integral ln(sinx) dx from x = 0 to ℿ/2. Now as sinx = sin(ℿ-x), then integral ln(sinx) dx from x = 0 to ℿ/2 = integral ln(sin(ℿ-x) dx from x = 0 to ℿ/2 Now, let u = ℿ-x. Then du = -dx. So, when x = 0, u = ℿ, and when x = ℿ/2, then u = ℿ/2. So, integral ln(sin(ℿ-x) dx from x = 0 to ℿ/2 = integral ln(sunu) (-du) from u = ℿ to ℿ/2 = Integral ln(sinu) du from u = ℿ/2 to ℿ = integral ln(sinx) dx from x = ℿ/2 to ℿ, when the dummy variable u is replaced by x So, integral ln(sinx) dx from x = 0 to ℿ/2 = integral ln(sinx) dx from x = ℿ/2 to ℿ. So, we don't need to rely on the graph to know these two definite integrals are equal.

Brilliant.

You guys hear that you learn by trying. The journey is the adventure

or just say that the graph of sin and cos in the interval (0, 2pi) are just flipped version of each other

Thank you

Thank you very much

Oh, that is really indegenious, I can never think out this on my own without hints!

@donegal79

5 жыл бұрын

performing integrals is ALL about experience

@mathcenter9481

4 жыл бұрын

Hlo

What if you used the complex definition of sine instead and then convert to polar coordinates for ln(a + bi) = ln(r) + iα

Can the same method be used if function's domain is < 0,π/2 ] ( 0 not included, improper integral !)

I realized I used to skip so many steps during my jee preparation

Couldn't this be solved using the complex definition of sine?

What if the same question was solved in indefinite integration? Please provide me solution.

@LetsSolveMathProblems

5 жыл бұрын

Indefinite integration does not yield an elementary solution for ln(sinx). I believe the solution for this particular case involves polylogarithm function, which I am personally not very well acquainted with.

very very useful

why is the area negative?, or should be negative since the curve is in that area?

@ethancheung1676

6 жыл бұрын

Zamir Coria because it is entirely below x-axis

@salomonmetre2117

3 жыл бұрын

@@ethancheung1676 Exactly. It is an algebraic area

Very nice

I have a question. ln[sin(x)] has a vertical asymptote at x=0. ln[sin(x)] ---> -inf as x ---> 0. How do you know that the original int of ln[sin(x)] from 0 to pi/2 converges in the first place? This method requires that the original integral converges to a finite number. If it doesn't, then your method just gives -inf = -inf. What am I missing?

@TheEternalVortex42

Жыл бұрын

It's true, to be rigorous you need to prove that it converges in the first place. One way is to write log(sin x) = log(x) + log(sin x / x). Since sin x / x -> 1 as x -> 0 the second term is bounded and it's easy to show that the improper integral of log(x) converges since it has a simple antiderivative.

@5:17 use angle addition identities

As long as this is an improper integral, is it competely justified to suppose it's finite? I would say no. 2I = I + c implies I=c only if we are sure I is finite. What you think?

@CHector1997

4 жыл бұрын

You have to prove that the improper integral exists and it´s a real number. To do that, you just need the comparison theorem(sin(x)/x tends to 1 as x tends to 0) and the fact of log is monotone. (improper integral of log(x) between 0 and 1 exists)

we can use the derivative of beta function here

My good sir can you please solve this using complex analysis?

really helpful

Just wondering, if you didn't have a picture of the graph to help you see the symmetry, is there a way to know that ln(sinx) is symmetric about pi/2?

@sarahgreenwood258

5 жыл бұрын

A sketch of t=sin(x) may help. A "better" way, but longer, would be to translate the curve by pi/2 units to the "right" and prove that this curve is even.

@user-pu4kj2sm7y

7 ай бұрын

Using f(pi-x)=f(x)

So I’m still learning integration from school and I’m still confused why he could just change the u to x and ignored the substitution part. Is it because after doing the substitution, the new integral becomes the same as the original integral so then the variable doesn’t matter anymore?

@primsiren1360

Жыл бұрын

use the substitution u = x. notice that the letter used doesn't matter in an integral

Used sinxcosx = 1/2sin(2x) and then had to prove that ln(sinx) is even at x=π/2 and got -π/2ln(2)

Isn’t the highlated area infinite though? Because the graph never touches the y-axis?

@LetsSolveMathProblems

6 жыл бұрын

Good question. Even if the region is not bounded, the area of the region can still be finite as long as the "region gets smaller quickly enough". For a one-dimensional version, consider the famous geometric series 1/2+1/4+1/8+1/16+.... This series evaluates to 1 (NOT infinity), even though we are adding an infinitely many values. If the values (or the areas) approach zero at a fast enough rate, the sum can approach a finite value.

@ianfowler9340

4 жыл бұрын

@@LetsSolveMathProblems "the area of the region can still be finite as long as the "region gets smaller quickly enough"." Yes, but how do we know that is the case in this example?

I tried to expand ln sin²x as -cos²x-cos⁴x/2-cos⁶x/3-... but was stuck at sum((2n-1)!!/(n(2n)!!)) until when I was falling asleep I suddenly thought of the same method ps I was calculating int arcsin x/x dx from 0 to 1 when I encountered this

Genius

why u consider ln (cos(u)) and ln(cos(x)) as same? suppose u have a function sin(x) and sin(2x) it would give same graph if u plot for x in first case and 2x in second case but you cannot say they are same if we take the same scale of x. and due to which their area would differe if we were to integrate them at same interval... would you please explain it.

@bmac3933

Жыл бұрын

2 years late but if you are still interested, when you make the substitution u = 2x, the bounds of the integral also change. So the areas are still the same since the change in bounds and the du = 2 dx effectively scale the area to make them equal.

Very interesting problem. Do you have other functions whose integration needs the treatment like this. Let me know if you have. I have some : ln(1+tanx) from 0 to pi/4 ; 1/(e^(sinx)+1) from 0 to 2pi ;

@oom_boudewijns6920

Жыл бұрын

are there ways to do these integrals without boundaries? If u are still here I would like to chat about it

What if you multiply by 1 and solve by parts. like how we would solve integral of ln(x)

@cjlagmay8611

5 жыл бұрын

You could use u = ln(x) and dv = dx. That way, you get du = 1/x dx and v = x

Wow this took me a while to get

@hilalbounebirat2091

5 жыл бұрын

It's 0

Here, I am thinking this as standard result.

can we use residue thm ?

Could you write sinx as an infinite product and then work it out?

@kevivmodi7019

3 жыл бұрын

the antiderivative is non elementary like the integral of e^(-x^2)

I like what u do and the way how u understand the things ... but this not the case for the teacher :I is there any other way to prove that integral of ln(sin(2x)) is equal to the double of I ?

@AymanSussy

5 жыл бұрын

I found this methode : integral from 0 to Pi of ln(sin(x)) ( witch is the same as the video show ) is I ( our integral ) + the integral from Pi/2 to Pi now we should prove that the second integral is equal to I as well ... using T=x - Pi/2 and cos(x) = sin(x+Pi/2) we got the result

The best

this qustn i saw in my board exam 12th standard @2018

Int-egg-ral lol The "e" in "integral" is just pronounced as a schwa: Int-uh-gral.

@GirishManjunathMusic

5 жыл бұрын

Intugrul, integral.

@practicalmediocrity3329

5 жыл бұрын

The American pronounciation is reslly vexing for me

@dothemath1554

4 жыл бұрын

Practical Mediocrity so is spelling...

@fashnek

4 жыл бұрын

Both pronunciations are regularly used.

stupid question but wouldn´t the complex form of Sinus help us? cause well it´s a stupid method of attack (but a stupid method of attack is better than none as my maths prof would say) but what makes this integral so nasty is the ln right? so a decent waay of getting rid of lns is using exponential functions right? And if we could insert the complex definition of sine in there shouldn´t we be able to create a perfectly integratable function? i´m very bad at log functions and properties cause i never learnt them so i got myself into a bind trying to get an integratable function but for someone a bit better at logarithms it should be easy right?

@hamzazaidi7042

6 жыл бұрын

Metalhammer1993 I tried to do it with the exopential method ,It came out to be pretty nasty

@Metalhammer1993

6 жыл бұрын

Hamzz Zaidi ahh okay so it doesn't work? Too bad fought there was an obvious solution to a difficult problem. But well you can't get rid of a sum within an ln right?

@SlingerDomb

6 жыл бұрын

that method using complex definition of sine is much much more harder which involves dilogarithm(i've never heard of it before until now) to solve this integral and also used many many substitutions. In case you want to see it www.integral-calculator.com/ plug the expression in and then go insane.

@Metalhammer1993

6 жыл бұрын

Zenith Juno thanks as i said i didn't know whether it works because my mathematical education is VERY lacking. I lit never had log properties in school and only saw them applied in university and by screwing up a log property i came with that stupid idea. I thought i could turn a sum inside the ln to a product which unfortunately doesn't quite work. If it worked we'd have been done in half a minute but too bad.

@SlingerDomb

6 жыл бұрын

I had that idea before but yeah unfortunately.

Como cuando sabes matemática pero no inglés :-)

@nelson6814

3 жыл бұрын

JAJAJAJAJAJ POR LO MENOS LE ENTENDÍ TODO LO QUE ESCRIBIÓ JAJA

Hi, I have a question. Why are you using lna+lab=lnab? Where that idea come from? I’ll appreciate if you can answer me

@nellvincervantes3223

4 жыл бұрын

Law of logarithm Ln(a) + ln(b) = ln(ab) Ln(a) - ln(b) = ln(a/b) Etc

@ursulau.7035

4 жыл бұрын

Nellvin Cervantes. Thanks, but I get that. The thing that I don’t understand is why he thought that will work

Why not just do the inverse integral of arcsin(e^x) from -inf to -0? Simple integral there and represents the same area

@dugong369

3 жыл бұрын

Good idea but the indefinite integral is non-elementary (per Wolfram) and the definite integral requires the same type of solution as the original problem.

This is in high school book of India... 😂

@user-uh9bo2im1h

10 ай бұрын

Well the difference between Indian school system is in Indian school system they will tell you if this do this in European university they teach you the basics and give you question like this

@SkullKnight2145

8 ай бұрын

The fact that this is true is even crazier😂😂😂😂

What if you don't have prior knowledge that the integrand plots a symmetrical graph?

@truppelito

6 жыл бұрын

Since sin(pi-x) = sin(x) you can avoid drawing a graph a sill reach that same conclusion

@SanjeevKumar-js4mu

5 жыл бұрын

Then use a property that Integral 0 to 2a f(x)dx= 2*Integral 0 to a f(x) dx Only if f(2a-x)=f(x)

@davidbrisbane7206

3 жыл бұрын

Show that: Integral ln(sinx) from x = 0 to ℿ/2 = integral ln(sinx) from x = ℿ/2 to ℿ. Now sin(ℿ-x) = sinx, as sin(ℿ-x) = sinℿ*cos(-x) + cosℿ*sin(-x) So, sin(ℿ-x) = 0*cos(-x) + (-1)*sin(-x) = 0 + (-1)*(-sinx) = sinx. Consider the integral ln(sinx) dx from x = 0 to ℿ/2. Now as sinx = sin(ℿ-x), then integral ln(sinx) dx from x = 0 to ℿ/2 = integral ln(sin(ℿ-x) dx from x = 0 to ℿ/2 Now, let u = ℿ-x. Then du = -dx. So, when x = 0, u = ℿ, and when x = ℿ/2, then u = ℿ/2. So, integral ln(sin(ℿ-x) dx from x = 0 to ℿ/2 = integral ln(sunu) (-du) from u = ℿ to ℿ/2 = Integral ln(sinu) du from u = ℿ/2 to ℿ = integral ln(sinx) dx from x = ℿ/2 to ℿ, when the dummy variable u is replaced by x So, integral ln(sinx) dx from x = 0 to ℿ/2 = integral ln(sinx) dx from x = ℿ/2 to ℿ. So, we don't need to rely on the graph to know these two definite integrals are equal.

Wait I thought that u = pi over 2 minus x so how can you. Just sub x for u?

@LetsSolveMathProblems

5 жыл бұрын

In definite integration, we can safely ignore the "relationship" between u and x after the substitution is complete (in fact, we can actually make the substitution x to pi/2-x if we wish!). To see what exactly happened, consider the three integrals: integral from 0 to 1 of x^2 dx, integral from 0 to 1 of u^2 du, and integral from 0 to 1 of (smiley face)^2 d(smiley face). All three integrals have the same value because definite integrals are independent of what letter is used to denote the variable (that is, if you change all the u's to x's or all the u's to (smiley face)'s, the value of the integral is unchanged). Perhaps the explanation in the video would be clearer if you think of it as making the u-substitution first THEN changing all the u's to x's to get an integral with the same value.

@MC-mx1mt

5 жыл бұрын

we have already corrected the bounds, so we can sub x for u now.

12:34 Sometimes, I have difficulty understanding how you don’t mix yourself in you writing all over the place

@donegal79

5 жыл бұрын

he does

dont you need to show the integral exist? both for x=0 and x=pi/2 the function integrated is not bounded there? Also the graph alone is not enough to justify..? You have to actually prove the integrals are equal i think. Cool video anyway and nice way of solving it!

it's sqrt(pi) / 2

Could we use Feynman's Technique?

@renerpho

6 жыл бұрын

I tried... But didn't quite get all the way to the answer. The most promising result I got by setting I(b) = integral_0^pi/2 ln(sin(x)*cos(b)) dx. Then we have I'(b) = -pi/2*tan(b), and I(b) = pi/2*ln(cos(b))+C. We are looking for I(0). Setting b = 0 gives I(0) = C, so we have I(b) = pi/2*ln(cos(b))+I(0). You can set b = pi/3 to get I(pi/3) = pi/2*ln(cos(pi/3))+I(0) = -pi/2*ln(2)+I(0). Since we want to prove that I(0) = -pi/2*ln(2), this reduces the problem to showing that I(pi/3) = 2*I(0). That looks promising. Unfortunately, it's where I get stuck. If you can show that integral_0^pi/2 ln(sin(x)/2) dx = 2*integral_0^pi/2 ln(sin(x)) dx, you are done. Good luck!

@donegal79

5 жыл бұрын

why dont you show us then? Name dropper

In preparation of engineering entrances in India, everyone has learnt dis as an std result to use in more complex problems lmao

@Bollibompa

5 жыл бұрын

That's sad. Especially since you're so subpar as actual engineers in the end.

@Bollibompa

5 жыл бұрын

@@sasmitvaidya3594 It does not produce good engineers so how is it good?

@_H_A_R_S_H_I_T_

5 жыл бұрын

@@Bollibompa India has the highest number of engineers in the world ...

@Bollibompa

5 жыл бұрын

@@_H_A_R_S_H_I_T_ Yeah, so? They have 1.3 billion people in the country, of course there will be many. Ever heard of quality over quantity? The Indian education system fosters not what is helpful for an engineer so the end result is mediocre or worse. No creativity, no Independence and no ingenuity.

@marvels3011

5 жыл бұрын

I agre with @bollibompa.I am an engineer myself in india.There are so many engineers(useless) who don't know even the basics of engineering and do rote learning to pass exams

what accent？

One time I was just laying on my bed when this integral suddenly jumped to my mind and I somehow quickly realized how to do it, I am not joking

That will be I/2 at last

2:15 sqrt(2pi)

@youkan4

5 жыл бұрын

no

I think this is a improper integral. U only use the idea of definite integral

It's spelled ìntegàl not íntègról

That is much easier than the equivalent 😝 [-∞ to 0] -∫ sin^-1(e^x) dx

@_H_A_R_S_H_I_T_

5 жыл бұрын

That's simple, just apply integration by parts...

@Ni999

5 жыл бұрын

@@_H_A_R_S_H_I_T_ Thanks, I already understand that works. Have you tried it and compared to the video? Of course, you're joking.

@lilyyy411

4 жыл бұрын

@@Ni999 its fun... you get i*(Li_2(1-e^(2iarcsin(e^x))--arcsin^2(e^x))from -infinity to 0

Redirecting to math world 🌍

7:34 "this x is NOT the same as that [above] x" .... 8:15 puts the "above" x and the "lower" x in a equation (therefore -wait for it- equating them) .. ??? Please more information needed still

@mesahusa

6 жыл бұрын

ThomasHaberkorn think of x as a arbitrary variable, and when he uses u-sub it becomes a “different” variable, but he “realigns” the equation back by modifying the bounds of integration so that it becomes the same result. An example would be if you integrated cos(x+pi) from 0 to pi, you could use u-sub u= x+ pi, as you can see it shifts the whole equation back by pi, so if you then shift the upper and lower bounds the other way, it cancels each other out and you’re left with essentially the same integral, and can treat u and x as the same thing.

@leo.y.comprendo

6 жыл бұрын

mesahusa Is this also valid for indefinite integrals?

@mesahusa

6 жыл бұрын

Keep in mind, indefinite integrals don't really mean anything in the real world if you're trying to actually find something. When your calculus professor is asking you to solve an indefinite integral, he wants you to give the most accurate answer, and bounds are just a part of it if you are given one. If not, back-substitution is the only way you can show you actually have the right answer.

just put it in the online integral calculator, humans will be obsolete one day anyway

Easy

This question is solved by class 12 students in India

@adnaniqbal2878

3 жыл бұрын

feeling proud on being a jee aspirant

J'ai du mal à imaginer que ce calcul, tout juste du niveau d'un élève de mpsi en milieu d'année, soit une trouvaille du "MIT" !!!!! C'est rigolo de voir que des connaissances assez banales peuvent être présentées de façon aussi "théâtrale" par des étudiants en manque de reconnaissance.

ln (sin(0)) actually is undefined Thus, I think there will always be a infinite small error

مافي بلعربي

@mohammedhubail1607

6 жыл бұрын

s.syrai sm ما في

@siphilipe

6 жыл бұрын

It was.

한국인이신가???

You cannot just substitute x=u-pi/2 and then rename u back to simply x. That is, generally spoken, simply wrong. I think you are using some kind of symmetry of cos(x) to pi/2...

@remysanlaville3085

6 жыл бұрын

why not? of course you can. you can change the name of the parameter to anything you want.

@matyourin

6 жыл бұрын

Because x is already used before with another value. Imagine I want to integrate cos(2x-2), then call u=2x-2 so it gets cos(u) and then call u=x so it gets cos(x)... that's not the same as before.

@hallig3_

6 жыл бұрын

When u change the variables u=2x-2 u also have to change the bounds and the dx to du wich ''corrects'' the fact that u renamed a variable according to the equation u = 2x-2

@SirFancyPants21

5 жыл бұрын

@@matyourin all he meant is that they are simply dummy variables. Yeah it makes it somewhat convoluted but it's not wrong.

Inteeeeegral..horroble

@LetsSolveMathProblems

6 жыл бұрын

Yes, I always thought it was pronounced "inˈteɡrəl" for some reason! I had a fun laugh a couple evenings ago when I realized I was totally wrong. Thank you for reminding me! I will strive to pronounce it "ˈin(t)əɡrəl" from now on. =)

@Buenofresser

6 жыл бұрын

LetsSolveMathProblems no problem :D keep up the good work

@ib9rt

6 жыл бұрын

The word "integral" has two different pronunciations in English. In the math form (noun) the stress is on the first syllable, but in the common language form (adjective) the stress is on the second syllable.

Oh so great Jesusbisnthe Son of God Believe in his sacrifice and resurrection and repent from sin

@darthsion3844

4 жыл бұрын

Relevance?

Not your finest explanation...at one point you were tripping over yourself. Amazing how you showed that sin x = cos(pi/2 -x)) but believed in another video that it was simply obvious that tangential circles bounded by a pair of external tangents have radii in a geometric ratio. ????? Confused.

horseshoe integration

a bit wordy

Too easy....

This integral is taught to 12th grade students in India

Please tell him someone.. if he sees a jee advanced question he would be just awestruck,If he looks this question as difficult.. it is an NCERT question broo😂😂😂😂

@diwakartiwari6496

3 жыл бұрын

Is there a proof in NCERT that ln(sinx) is symmetrical about π/2 ? This question is not as easy one if you include the proof part. I have also solved this problem and used it as a standard result,but never thought of proving it,nor could have I.

@sakshampaliwal8170

3 жыл бұрын

@@diwakartiwari6496 then you must start focusing on studying the basics, because this proof is not a great thing....

@diwakartiwari6496

3 жыл бұрын

@@sakshampaliwal8170 It may not be a great thing but also not very intuitive. Talking about the NCERT part: They didn't appraise the symmetry part which is the beauty of this problem and must be touched while writing solution of this problem. About jee: This is the very first cause of declining creativity in millions of students who just want to be engineers but don't feel an engineer inside them. As a result they end up in frustration; and start preparing for SSC or UPSC or railway xams. Even jee advanced can be cracked on the grounds of ratta maar. What they solve the books are previous year questions of western countries' competitions. I have a big collection of questions from cengage, arihant,rd Sharma and many more which are exactly copied from western writer's books. You may be a bright student but the way you addressed him in ur main comment was immature and ostentatious. It will never boost ur or ur nation's image,but people get it as if you are boasting off.

LAMAOO They teach this stuff way before undergrads here in India.

Thank you very much