@Rohan_Shinde1 can you share some of the book names that you used for studying these so called "beyond JEE scope" concepts? I'm interested in learning these after May. Especially calculus and algebra.

# Thread to share Good Questions from Calculus which requires deep thinking

**Rohan_Shinde1**#307

@Supreeta_Sen I specially love calculus, and in particular sequence and series along with Definite integrals.

I prefer to read the following books

1)Inside interesting integrals by Paul J. Nahin.

2)Irresistible integrals by Boris and Moll

3)Complex Numbers by Titu Andreescu

4)A synopsis on Pure and Elementary Mathematics by G. S. Carr

5)Calculus Volume 1 and 2 by Tom Apostol, and by the way Michael Spivak is good too for calculus

6) A treatise on Integral calculus Volume 1 and 2 by Joseph Edwards.

7) Art of Problem solving by Paul Zeitz

8)Problem Solving Strategies by Arthur Engel(Didnt read quite much). Also Mathematical Circles(Russian Experience) is also good.

9)Geometry Revisited by H. S. M Coxeter

10)Euclidean Geometry for Olympiads by Evan Chen.

Also solving some Putnam Exam papers as well as HMMT problems along with problems from magazines like Crux mathematicorum and American Mathematical monthly are also helpful. Also try mathematics and science forums and websites like Math stackexchange , Physics stackexchange, Brilliant.org

**Rohan_Shinde1**#309

@Mayank_Chowdhary

It is quite obvious that x_k are the positive roots of the equation \tan x= x

Using some coordinate geometry we can find that the locus of points of contact is

x^2-y^2=x^2y^2

Using this information we see that we need to find \sum_{k=1}^{\infty} \frac {x_k^2+1}{x_k^2(x_k^2+2)}=\frac 12\left( \sum_{k=1}^{\infty} \frac {1}{x_k^2} +\sum_{k=1}^{\infty} \frac {1}{x_k^2 +2}\right)

Now writing power series of

\frac {\sin x-x\cos x}{x^3}

about 0 will give a polynomial with positive roots of \tan x=x.

Using Vieta's formula we see that

\sum_{k=1}^{\infty} \frac {1}{x_k^2}=\frac {\text{ -Coefficient of x²}}{\text {Constant term}}=\frac {1}{10}

Also changing x\rightarrow \sqrt {x-2} in the power series and using the same method above gives

\sum_{k=1}^{\infty} \frac {1}{x_k^2+2}=\frac {\text{ -Coefficient of x}}{\text {Constant term}}=\frac {5\sqrt 2\sinh(\sqrt 2)−6\cosh(\sqrt 2)}{4(2\cosh(\sqrt 2)−\sqrt 2\sinh(\sqrt 2))}

**arush_2019**#310

Oops I forgot that \int _0 ^{π \over 2} cosx = -{π \over 2 } × ln2

So did with Gamma

\int _0 ^{π \over 2} cosx = {1 \over 2} ln(1-sin^2x)

On expanding..

=-{1 \over 2}(sin^2x + \frac{sin^4x}{2} + \frac{sin^6x}{3}.......)

So by using gamma function

= -{1 \over 2}× \gamma \frac{ \frac {( 2+1)}{2} \gamma {1 \over 2 }}{ 2× \gamma { \frac {2+2}{2}}}

+\gamma \frac{ \frac {( 4+1)}{2} \gamma {1 \over 2 }}{2×{ \gamma \frac {4+2}{2}}}.......

On taking common and converting got a series..

- \frac{π}{4} × \sum _1 ^{ \infty } \frac{1.3.5.7....({2r-1})}{r.r! × 2^r }

So can anyone prove that this series also comes to be 2ln2

**arush_2019**#312

@Rohan_Shinde1

Am I doing wrong anywhere ????

So why sum is not coming and it is diverging???

Why my gamma doesn't shows like your gamma???

I too write \gamma☺

..

It's very hard to do such latex ..

I missed out several brackets and at last time out so cannot make further changes

..

**Rohan_Shinde1**#317

Yep I do. And that to the extent that my MSE and Instagram user names are Digamma and Digamma_Integrator. Am not I a little weird?

**Rohan_Shinde1**#320

Hints:

I=\int \frac {x^2\cos^2x -x\sin x-\cos x-1}{(1+x\sin x)^2}dx

=\int \left(-1+ \frac {x^2+x\sin x-\cos x}{(1+\sin x)^2}\right) dx

Now for the second part I would like to know what is \frac {d}{dx}\left(\frac {x\cos x}{1+\sin x}\right)

Gotcha got something?

As for the answer the answer according to me is amazingly 0 as per my calculation

**Azimuddin_2019**#321

Awesome !! @Rohan_Shinde1 bro , between can u tell how u think like this the derivative will be ??

**Rohan_Shinde1**#324

Actually the square of function reminded me of the quotient rule of differentiation, the x^2 term in numerator of our integrand gave me a thought that the antiderivative must have have x multiplied with some other function and working out with different trig functions, the antiderivative just popped before me.

**Azimuddin_2019**#326

Yeah nice I got it , thx a lot @Rohan_Shinde1 bro , lol likely that friend is common to us @Naman bro