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

#429

See (–1)^(2x) = 1 is only true until we know that x= natural number, here we can't directly say that f(t) = x for some natural number

#430

Oh yes didn't think about the non integer values. Sorry I will try the question again.

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#431

Oh yes got it

Just write (-1)=e^{i\pi}

In the integral and then use e^{ix}=\cos x+i\sin x

After that the real part will vanish like demonstrated above by showing some J=-J and only the imaginary unit will be left.

@Tushar_Rathore

Bring on the next one.

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#432
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#433

Is wale question ka answer mujhe bhi nhi pta

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#434

I also did as did by Azimuddin Sheikh so did not post my solution as would be repeated but I just have a thought that since zeta function is involved -\ln x=t would do some work. But I have not tried that substitution yet so I will try that next morning.

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#435

@Tushar_Rathore
it is 1
It is easily converting..

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#436

Answer is not known to me. I will confirm it tomorrow bro
Let see what @Rohan_Shinde1 gets

#437

I get answer as 1. @Tushar_Rathore

Ends up wrapping to
\displaystyle \sum_{n=1}^{\infty} \left(\frac {1}{(n-1)!}-\frac {1}{n!}\right)=e-(e-1)=1

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#438

Question

Consider a sequence \{F_n\}_{n\ge 0}
( Which is basically the Fibonacci sequence) such that F_0=0 and F_1=1 whereas F_{n+2}=F_{n+1}+F_n

Then prove that
\displaystyle \sum_{n\ge 1} \arctan \left(\frac {(-1)^{n+1}}{F_{n+1}(F_n+F_{n+2})}\right) =\arctan (\sqrt 5 -2)

Cheers... Good night everyone..

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#441

Lol only question on this thread that i was able to do under 30sec

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#443

\int_0^{\pi/2}\tan^{-1}(\sin x) \ dx.

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#444
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#445

Sinx=tan@ might work🤔

#446

Not working.
I don't think it has elementary antiderivative

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#447

It has @pratyaksh_tyagi bro. Hint : integration under differentiation. Also try this one too: Find value of t for which the expression f(t)=\int^{4}_{0}\bigg|4x-x^2-t\bigg|dx is minimum.

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#448

It's very lengthy.
I(a) = \int_{0}^{\frac{\pi}{2}} tan^{-1} ( asinx) dx
Then differentiate.

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#449

pratyaksh_tyagi is the final answer correct from yours method

1/\sqrt{2} *2 (\log({ sqrt 2 -1)/sqrt 2 +1) }

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#450

@Azimuddin_Sheikh 4?

#451

Ignore post #450 that's a wrong answer