A gem of an integral from the JEE advanced calculus exam

Here's my first integral from the Indian JEE advanced calculus exam. Solution development involved using some nice trigonometric tricks and integration by parts.

[kewalmer]

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.

[LeonidRachevsky]

*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 + 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 +
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. - - 1/sqrt(1+(pi/4)^2))_
10. + + 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)_*

[BREAD-kczp]

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

[anonymous_]

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.

[MrWael]

Very nice and smart simplification. Really Talented.

[bakbasbg]

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:

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 😊

[fengshengqin]

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 .

[UzziWallendorf]

->xsin/xcos ²)dx pie/4. 0->4/pie pie/4 ⁰

[wryanihad]

Complex method there is an easer way to solve which is. By part from begin

[skyethebi]

0:50 my algebra 1 teacher called that a FFOO (fancy form of one) which he pronounced foo-foo

[adityavsx]

i am prepping for jee and believe me, your videos give so many more tricks and intuition. thankyou so much

[rk]

You could sell a catalog of your bag of tricks for big bucks!

[Anonymous-Indian..]

These are my old
I remembered this question was in by JEE book
I also gave JEE Advanced
And succeeded in that mission

[dhruv]

i am going to give adv in next week hoping for best

[ガアラ-hh]

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

[theblainefarm]

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.

[amoghdadhich]

No way I'd have enough patience or skill to do this in an exam though 🙃

[vikrantsingh]

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

[zunaidparker]

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.

[shanmugasundaram]

Very nice trick.The cancellation of the second integral is very interesting.