Thread to share Good Questions from Calculus which requires deep thinking

#409

bro haven't looked in b/w posts .
@Rohan_Shinde1

#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}

1 Like
#411

2nd is Feynman trick write x^b And differentiate.
3rd is very famous aka "dirichlet integral". Proof is in this thread

3 Likes
#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.

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#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

2 Likes
#414

Anyone done this @Rohan_Shinde1, @Siddhant_Mudholkar , @arush_kumar_singh , @Yash_Srivastava2 , @pratyaksh_tyagi
#404

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#415

Getting 1 only broπŸ€” dunno what's wrong @Tushar_Rathore


see #404 @Azimuddin_Sheikh

#416

@Tushar_Rathore

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

5 Likes
#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

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#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.

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#419

Yaa sure why not , wait for a min

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#420

Here is your first question

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#421

0?

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#422

@Tushar_Rathore @Rohan_Shinde1
Post questions of other topics too instead of integral if you have (good ques)...
Aaj integral se pet bhar gayaπŸ˜…

2 Likes
#423

Nope its iota (i)

#424

@Rohan_Shinde1
Try this one also
The first integral


If you have diffrent approach please do share

#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?

2 Likes
#426

(–1)^(x) = –(–1)^(–x)

#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

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#428

@Rohan_Shinde1 You are not entirely right ...
What if x=1/2

1 Like