E Power Eye Pie Plus One

Calculus Level 3

If the following integral equals to $Z$

$\large \displaystyle \int_0^{\pi /2} \left ( (\sin x)^{\cos x} - (\cos x)^{\sin x} \right) \mathrm{d}x$

What is the value of $\lfloor 1000Z \rfloor$ ?

The answer is 0.

1 solution

Pranav Arora
May 21, 2014

LOL....what was that? :D

As for the solution, notice that the integral is:

$\displaystyle \int_0^{\pi/2} (\sin x)^{\cos x}\,dx-\int_0^{\pi/2} (\cos x)^{\sin x}\,dx$

The second integral is same as the first one and this can be shown by using

$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$

Hence, the integrals cancel and the answer is $0$ .

He has given the answer in the title! LOL

Anish Puthuraya - 7 years ago

:D that is funny !! e^{ i \pi +1 } is answer

Mohit Maheshwari - 7 years ago

This is why I never like Brilliant. I've just posted my solution here & comment but they delete them all.

Anastasiya Romanova - 7 years ago

To clarify what happened, your solution simply stated

What a joke! It's zero by symmetry. Try this one (click the problem):

$\int_0^{\Large\frac{\pi}{4}}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\, \exp\left[\frac{x^2-1}{x^2+1}\right]\, dx$ .

It was not a useful solution as it did not contain any explanations. Furthermore, the linked problem did not help explain and would merely introduce cofusion, which is why I deleted it.

Calvin Lin Staff - 6 years, 5 months ago

E Power Eye Pie Plus One

The answer is 0.

1 solution

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