This little brain teaser should neither be too easy nor too difficult for you; in other words, it is just right for all of you!

Define the following function f(x), where x is a real number between -1 and 1 both inclusive, as:

f(x) = (x / 2) * sqrt(1 - x^2) + the integral between x and 1 of sqrt(1 - t^2) with respect to t.

sqrt(x) returns the positive square root of x.

Now let the function Mystery be defined as follows:

If x is between 0 and pi, then Mystery(x) is the unique number between -1 and 1 both inclusive such that f(Mystery(x)) = x / 2.

The question is: which familiar function is the Mystery function above mimicking exactly?

I've seen that integral and and its polar coordinate version (sin^2 x) so many times...

I'll do the integral itself first, and then compute the answer.

f(x) = (x / 2) * sqrt(1 - x^2) - integral from 1 to x of sqrt(1 - t^2) dt (because all the cool people put their inputs at the top of the integral)

int() denotes an indefinite integral.
sin^n() means (sin())^n.

t -> cos u
dt -> - sin u du
-int(sqrt(1 - t^2))dt =
= int(sqrt(1 - cos^2(u)) sin u du
= int(sin^2(u) du)
= -sin u cos u + int(cos^2(u)du)
= -sin u cos u + int((1 - sin^2(u))du)
= -sin u cos u + u - int(sin^2(u)du)
(move the integral to the left side because OMG they're the same; divide by 2)
= 1 / 2 (u - sin u cos u)
= 1 / 2 (arccos t - t sqrt(1 - t^2))

So the formula for f becomes
f(x) = x / 2 sqrt(1 - x^2) - x / 2 sqrt(1 - x^2) + 1 / 2 arccos x - 1 / 2 arccos 1 + 0
f(x) = 1 / 2 arccos x
Let Mystery(x) = cos(x). Then f(Mystery(x)) = x / 2 as required!

Good answer. The Mystery function is indeed mimicking the cosine function.

I'm only saying 'mimicking' rather than 'is equal to' because the cosine function is defined throughout the whole real line, whereas the Mystery function above is only defined from x = 0 to x = pi. That can be rectified, however, by adding:

If pi < x < 2pi, then Mystery(x) = Mystery(2pi - x)
If x = 2n pi + a, where 0<=a<2pi and n is any integer, then Mystery(x) = Mystery(a).

If the above are added to the definition of Mystery, then the Mystery becomes equal to the cosine function exactly.