bro haven't looked in b/w posts .

@Rohan_Shinde1

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

**Prashant3**#410

\int_{0}^{\pi/2}\frac{x}{tan(x)}dx share your approach

\int_{0}^{1}\frac{x^2-1}{log(x)}dx

\int_{0}^{\infty}\frac{sin(x)}{x}

**pratyaksh_2019**#411

2nd is Feynman trick write x^b And differentiate.

3rd is very famous aka "dirichlet integral". Proof is in this thread

**Prashant3**#412

yes bro first one too possible by differentiation under integral.

for first writing x as \arctan(a tan(x))

but for third integral is there any simple method.

**arush_2019**#413

For third one

\int _0^{\infty} \frac {sinx}{x} = \int _0^1 \frac {sinx}{x} +\int _1^{\infty} \frac {sinx}{x}

For second integral put x \to {1 \over x }

=\int _0^1 \frac 1x Γ {sin{ \frac 1x}}

So integral is

\int _0^1 \frac {sinx}{x} + \int _0^1 \frac 1x Γ {sin{ \frac 1x}}

Now just use expansions of sinx

**Tushar_2019**#414

Anyone done this @Rohan_Shinde1, @Siddhant_Mudholkar , @arush_kumar_singh , @Yash_Srivastava2 , @pratyaksh_tyagi

#404

**Siddhant_2019**#415

Getting 1 only broπ€ dunno what's wrong @Tushar_Rathore

see #404 @Azimuddin_Sheikh

**Rohan_Shinde1**#416

@Tushar_Rathore

I have asked it on MSE you may refer there my try as well as some other methods

**Tushar_2019**#417

Actually that was the tricky part(|cos|) in which i also confused

Anyways this question was maked today by my brother. He have more marvellous questions if u want i can share @Rohan_Shinde1

**Rohan_Shinde1**#418

Yes it would be my pleasure to devour them. Bring them on buddy @Tushar_Rathore. Can you post it right now, I wanted to see them at earliest.

**arush_2019**#422

@Tushar_Rathore @Rohan_Shinde1

Post questions of other topics too instead of integral if you have (good ques)...

Aaj integral se pet bhar gayaπ

**arush_2019**#424

@Rohan_Shinde1

Try this one also

The first integral

If you have diffrent approach please do share

**Rohan_Shinde1**#425

How can this be possible?

Let I=\int_a^b (-1)^{f(t)} dt

=\int_a^b (-1)^{f(a+b-t)} dt

=\int_a^b (-1)^{1-f(t)} dt

= -\int_a^b (-1)^{-f(t)}dt

But (-1)^x=(-1)^{-x}

So I=-\int_a^b (-1)^{f(t)}dt= -I

Hence I=0 so how is the answer i?

**Rohan_Shinde1**#427

@Tushar_Rathore You mean to say for example

(-1)^3=-(-1)^{-3}

\Rightarrow -1=-\frac {1}{(-1)^3}

\Rightarrow -1=-(-1)

\Rightarrow -1=1