Here's my first integral from the Indian JEE advanced calculus exam. Solution development involved using some nice trigonometric tricks and integration by parts.
hey, I got an alternate solution: write the integral as xcosx/(xsinx+cos)^2. xsecx : now xcosx is the derivative of xsinx+cosx now just apply by parts and you should be able to solve it instantly.
*11 STEPS:* 1. multiply and divide _x^2/(x*sin(x) + cos(x))^2_ by _(1+x^2)_ 2 "inject" _1/(1+x^2)_ into the denominator, so the integral will look as follows: *_integral(0, pi/4) of x^2/((1+x^2)*(sin(x)*x/(sqrt(1+x^2) + cos(x)*1/(sqrt(1+x^2))^2_* 3. notice that the absolute value of _x/sqrt(1+x^2)_ and _1/sqrt(1+x^2)_ does not exceed 1 and the sum of the squares of these terms equals 1 4. therefore, they can be "treated" as *_cos(u)_* and *_sin(u)_* respectively, where *_u_** = **_arcsin(1/sqrt(1+x^2))_* , so we get *_integral(0, pi/4) of x^2/((1+x^2)*(sin(x)*cos(arcsin(1/(sqrt(1+x^2))) + cos(x)*sin(arcsin(1/sqrt(1+x^2)))^2_* 5. apply *sin of the sum* formula and get: *integral(0, pi/4) of x^2/((1+x^2)*(sin(x + arcsin(1/sqrt(1+x^2))))^2* 6. notice *d(x+ arcsin(1/sqrt(1+x^2))) = x^2/(1+x^2)dx* , so we get *_integral(0, pi/4) of d(x+ arcsin(1/sqrt(1+x^2)))/sin(x+ arcsin(1/sqrt(1+x^2)))^2_* , which, in fact, equals to *_-cot(x+ arcsin(1/sqrt(1+x^2)))_* 7. _cot(0 + arcsin(0+1)) = cot(0+pi/2) = 0_ 8. compute _cos_ and _sin_ for _cot(pi/4 + arcsin(pi/4 + 1/sqrt(1+(pi/4)^2)))_ separately keeping in mind _cos(pi/4) = sin(pi/4) = 1/sqrt(2)_ 9. _cos(pi/4)*cos(arcsin(1/sqrt(1+(pi/4)^2))) - sin(p/4)*sin(arcsin(1/sqrt(1+(pi/4)^2)))=cos(pi/4)*((pi/4)/sqrt(1+(pi/4)^2) - 1/sqrt(1+(pi/4)^2))_ 10. _sin(pi/4)*cos(arcsin(1/sqrt(1+(pi/4)^2))) + cos(p/4)*sin(arcsin(1/sqrt(1+(pi/4)^2)))=cos(pi/4)*((pi/4)/sqrt(1+(pi/4)^2) + 1/sqrt(1+(pi/4)^2))_ 11. finally we get *_-cot(pi/4 + arcsin(pi/4 + 1/sqrt(1+(pi/4)^2))) = -(pi/4 - 1)/(pi/4+1)=(4-pi)/(4+pi)_*
Another way : write the integrand as [(xsec(x)).xcos(x)]/(xsinx + cosx)^2 then integrate by parts , (x secx) as 1 st function and the rest one as 2 nd function. in that way it will be much easier and straightforward... no cancelation is required.
Yes I have calculated it this way some time ago but his way is a little bit more general but a little bit easier would be to calculate by parts I_{2} Your proposition can be found in my comment on one of the Steve's video (blackpenredpen) Later Steve used this approach in his video
Thanks for noticing! Been watching your videos for quite some time to look at integrals differently and discover new ways of thinking, instead of what they usually teach us when preparing for advanced
The first sight of this integral is to induce tanx/2 ,then do it in my mind ,NOK. The second thought is the differential law of division ，and creat an function f(x) ,let f(x)*(xsinx+cosx)'-f'(x)*(xsinx+cosx)=x^2 .
That's Exactly how I did it , I solved the last differential equation by multiplying x² on the right side with (sin²(x)+cos²(x)) and then compared the coefficients of sin(x) and cos(x) on both side of equation. This gives f(x)=sin(x)-x*cos(x)
A bit late to the party but I have got an even easier and faster way to solve this Substitute x=tan(θ) Hence dx=sec^2(θ) dθ After substituting, multiply numerator and denominator by cos^2(θ) The denominator will become: (Sin(θ)sin(tanθ)+cos(θ)cos(tanθ))^2 Which simplifies to (cos(θ-tanθ))^2 Substitute tanθ - θ= u Hence du=dθ(sec^2θ -1)=tan^2θ dθ which is exactly what the numerator is Hence we get integral sec^2(u) du Which is pretty basic and can now be easily solved Btw I am going to give this monster of an exam in few months, wish me luck 😊
I wouldn't say that these are easy problems but doesn't require much of intuition rather focuses on how much of tricks or manipulation you can do with what you have been provided , but if you want to solve some really great problems , I would recommend you to solve problems from CMI(Chennai Mathematical Institute , India) exams , specifically from their part B of Question paper 🙃🙃 I hope you consider my recommendation their question wouldn't disappoint you for Both Undergraduation and Post Graduation
Nice! I have seen a lot of different methods to solve the same integral, but yours is quite clever. I actually developed a new method for integration by parts that uses a reversal of the quotient rule then applied it to this integral.
How the hell can someone solve this how do I even start to learn this? Okay I’m in grade 10 being able to use u-Substitutin Partial Integration but thats it
Oh dont worry. This integral is from the JEE exams in India. The integrals there aren't really tough they just need some tricks. There are thousands of coaching centres and tutors in india that teach tricks specifically suitable for solving these type of integrals. If you really want to learn integration, solve as many integrals as possible. Sometimes even terrifying integrals can be solved using u subs and integration by parts.
I'm more interested in knowing how to approach this from first principles. These clever substitutions are clearly shortcuts leveraging foreknowledge what the answer is, and don't teach any integration techniques. My first instinct is to substitute x-->pi/2-x and try from there. I'm not sure if that would work.
Actually this question was given in the Jee mains examination which is an exam u have to clear to be elligible for the Jee advanced which is even harder, i would suggest u to solve the hardest jee adv. Question,it is of limit as a sum topic from definite integrals,keep in mind that this exam is for 12 th graders.
can you solve the jee advanced questions from 2016 paper 2? it was a mixed question of limits and integration and considered the toughest question till date.
hey, I got an alternate solution: write the integral as xcosx/(xsinx+cos)^2. xsecx : now xcosx is the derivative of xsinx+cosx now just apply by parts and you should be able to solve it instantly.
I did the same way and it came out really nicely
Same this method is given in my JEE book
I am sorry if I am making a mistake, but isn't the derivative of xsinx + cosx = xcosx-sinx. Can you tell me what I am doing wrong here??
Yeah. I got that now. Thanks for the answer tho. My stupid brain was taking x to be a constant and just getting the wrong answer.
@@Anonymous-Indian..2003 har jagah iit naam likhna jaruri h?
*11 STEPS:*
1. multiply and divide _x^2/(x*sin(x) + cos(x))^2_ by _(1+x^2)_
2 "inject" _1/(1+x^2)_ into the denominator, so the integral will look as follows:
*_integral(0, pi/4) of x^2/((1+x^2)*(sin(x)*x/(sqrt(1+x^2) + cos(x)*1/(sqrt(1+x^2))^2_*
3. notice that the absolute value of _x/sqrt(1+x^2)_ and _1/sqrt(1+x^2)_ does not exceed 1 and the sum of the squares of these terms equals 1
4. therefore, they can be "treated" as *_cos(u)_* and *_sin(u)_* respectively, where *_u_** = **_arcsin(1/sqrt(1+x^2))_* , so we get
*_integral(0, pi/4) of x^2/((1+x^2)*(sin(x)*cos(arcsin(1/(sqrt(1+x^2))) + cos(x)*sin(arcsin(1/sqrt(1+x^2)))^2_*
5. apply *sin of the sum* formula and get: *integral(0, pi/4) of x^2/((1+x^2)*(sin(x + arcsin(1/sqrt(1+x^2))))^2*
6. notice *d(x+ arcsin(1/sqrt(1+x^2))) = x^2/(1+x^2)dx* , so we get *_integral(0, pi/4) of d(x+ arcsin(1/sqrt(1+x^2)))/sin(x+ arcsin(1/sqrt(1+x^2)))^2_* ,
which, in fact, equals to *_-cot(x+ arcsin(1/sqrt(1+x^2)))_*
7. _cot(0 + arcsin(0+1)) = cot(0+pi/2) = 0_
8. compute _cos_ and _sin_ for _cot(pi/4 + arcsin(pi/4 + 1/sqrt(1+(pi/4)^2)))_ separately keeping in mind _cos(pi/4) = sin(pi/4) = 1/sqrt(2)_
9. _cos(pi/4)*cos(arcsin(1/sqrt(1+(pi/4)^2))) - sin(p/4)*sin(arcsin(1/sqrt(1+(pi/4)^2)))=cos(pi/4)*((pi/4)/sqrt(1+(pi/4)^2) - 1/sqrt(1+(pi/4)^2))_
10. _sin(pi/4)*cos(arcsin(1/sqrt(1+(pi/4)^2))) + cos(p/4)*sin(arcsin(1/sqrt(1+(pi/4)^2)))=cos(pi/4)*((pi/4)/sqrt(1+(pi/4)^2) + 1/sqrt(1+(pi/4)^2))_
11. finally we get *_-cot(pi/4 + arcsin(pi/4 + 1/sqrt(1+(pi/4)^2))) = -(pi/4 - 1)/(pi/4+1)=(4-pi)/(4+pi)_*
I used same method
Another way :
write the integrand as [(xsec(x)).xcos(x)]/(xsinx + cosx)^2
then integrate by parts , (x secx) as 1 st function and the rest one as 2 nd function. in that way it will be much easier and straightforward... no cancelation is required.
Yes I have calculated it this way some time ago but his way is a little bit more general
but a little bit easier would be to calculate by parts I_{2}
Your proposition can be found in my comment on one of the Steve's video (blackpenredpen)
Later Steve used this approach in his video
Thanks for your innovative solution.
Thanks for noticing! Been watching your videos for quite some time to look at integrals differently and discover new ways of thinking, instead of what they usually teach us when preparing for advanced
The first sight of this integral is to induce tanx/2 ,then do it in my mind ,NOK. The second thought is the differential law of division ，and creat an function f(x) ,let f(x)*(xsinx+cosx)'-f'(x)*(xsinx+cosx)=x^2 .
That's Exactly how I did it , I solved the last differential equation by multiplying x² on the right side with (sin²(x)+cos²(x)) and then compared the coefficients of sin(x) and cos(x) on both side of equation. This gives f(x)=sin(x)-x*cos(x)
Very nice and smart simplification. Really Talented.
A bit late to the party but I have got an even easier and faster way to solve this
Substitute x=tan(θ)
Hence dx=sec^2(θ) dθ
After substituting, multiply numerator and denominator by cos^2(θ)
The denominator will become:
(Sin(θ)sin(tanθ)+cos(θ)cos(tanθ))^2
Which simplifies to (cos(θ-tanθ))^2
Substitute tanθ - θ= u
Hence du=dθ(sec^2θ -1)=tan^2θ dθ which is exactly what the numerator is
Hence we get integral sec^2(u) du
Which is pretty basic and can now be easily solved
Btw I am going to give this monster of an exam in few months, wish me luck 😊
Hiii did you qualify? Btw thanks for the genius sln❤
->xsin/xcos ²)dx pie/4. 0->4/pie pie/4 ⁰
Complex method there is an easer way to solve which is. By part from begin
You could sell a catalog of your bag of tricks for big bucks!
0:50 my algebra 1 teacher called that a FFOO (fancy form of one) which he pronounced foo-foo
If only he called a fancy form of zero a “foo-foz”
I wouldn't say that these are easy problems but doesn't require much of intuition rather focuses on how much of tricks or manipulation you can do with what you have been provided , but if you want to solve some really great problems , I would recommend you to solve problems from CMI(Chennai Mathematical Institute , India) exams , specifically from their part B of Question paper
🙃🙃 I hope you consider my recommendation their question wouldn't disappoint you for Both Undergraduation and Post Graduation
Or ISI exam
Nice! I have seen a lot of different methods to solve the same integral, but yours is quite clever. I actually developed a new method for integration by parts that uses a reversal of the quotient rule then applied it to this integral.
i am prepping for jee and believe me, your videos give so many more tricks and intuition. thankyou so much
Very nice intergral :)
How the hell can someone solve this how do I even start to learn this? Okay I’m in grade 10 being able to use u-Substitutin Partial Integration but thats it
Oh dont worry. This integral is from the JEE exams in India. The integrals there aren't really tough they just need some tricks. There are thousands of coaching centres and tutors in india that teach tricks specifically suitable for solving these type of integrals. If you really want to learn integration, solve as many integrals as possible. Sometimes even terrifying integrals can be solved using u subs and integration by parts.
@@maths_505but we have to find the tricks
These are my old memories........
I remembered this question was in by JEE book
I also gave JEE Advanced 2021........
And succeeded in that mission
You are 20 years old
@@maalikserebryakov ok?
@@maalikserebryakov I mean PPL usually give jee when theyre 17 and start prep from 15 so yeah it's kinda nostalogic
Yeah....
But I was a dropper......
I start my preparation after class 12th when I was 17 and qualified JEE when I was 18......
@@Anonymous-Indian..2003 dont worry you are smart he is not lol dumb people always need to make rude comments in comment section of videos
nice one...
i am going to give adv in next week hoping for best
All the best bro.......❤❤❤
I can feel your emotion.......
Two years ago, I was in same condition......
Lol would be funny af to watch all these integration videos and come out with a fail 🤣🤣🤣🤣
with that said Good luck bro
@@Anonymous-Indian..2003thanks bro do you qualify?
@@dhruv9657
See description for motivation
No way I'd have enough patience or skill to do this in an exam though 🙃
Surprise! cancellation!! BIGGER SURPRISE! I DID IT WRONG AND EVERYTHING CANCELLED, and I got a niche proof that zero does indeed equates to 0
😂😂😂
Any other alternate soln???
I'm more interested in knowing how to approach this from first principles. These clever substitutions are clearly shortcuts leveraging foreknowledge what the answer is, and don't teach any integration techniques.
My first instinct is to substitute x-->pi/2-x and try from there. I'm not sure if that would work.
Oh believe me it would but that would be extremely painful🤣
Bro trust me you don' wanna go down that road🤣🤣🤣
@@maths_505 how would you then do this problem in a more natural way without using "foreknowledge" substitutions?
An off topic question- Being an asian, Is that your natural accent?
Actually you've to do this in
Very nice trick.The cancellation of the second integral is very interesting.
Preparations for integration by parts nice but I would integrate by parts I_{2}
Actually this question was given in the Jee mains examination which is an exam u have to clear to be elligible for the Jee advanced which is even harder, i would suggest u to solve the hardest jee adv. Question,it is of limit as a sum topic from definite integrals,keep in mind that this exam is for 12 th graders.
I think if you do the integral by making it xcosx.xsecx dx/(xsinx+cosx)^2 it comes out without much complications
Yeah.....
Same solution is given in JEE books
It is also easy to spot
I am studying in class 10 ❤
the moment i saw xcosx+sinx ^2 i knew it's a differentiation lmao
can you solve the jee advanced questions from 2016 paper 2? it was a mixed question of limits and integration and considered the toughest question till date.
Do this one integral((sec²x)/(secx + tanx)^9/2)
It's from the JEE Adv 2012
secx+tanx=u
secx-tanx=1/u
secx=(u+1/u)/2
Now the integral becomes
[(u+1/u)/2u^(11/2)]du
Or something like this
Correct or not I don't know
Please solve jee main questions
demoniator: HARMONIC ADITION