Prove that the function f(x)=arcsinx+3arccosx+arcsin(2x*square root(1-x^2)) is constant.

We could use derivative of the function to prove that f(x)
is constant. If we'll prove that the derivative of f(x) is cancelling, therefore the
function is constant, since the derivative of any constant function is
0.

We'll differentiate the function with respect to
x:

f'(x) = 1/sqrt(1-x^2) - 3/sqrt(1-x^2) +
[2x*sqrt(1-x)^2]'/sqrt[1-4x^2*(1-x^2)]

f'(x) =
-2/sqrt(1-x^2) + [2sqrt(1-x^2) - 2x*2x/2sqrt(1-x^2)]/sqrt(1-4x^2 +
4x^4)

f'(x) = -2/sqrt(1-x^2) +
[2(1-x^2)-2x^2]/(1-2x^2)sqrt(1-x^2)

f'(x) = -2/sqrt(1-x^2)
+ (2-4x^2)/(1-2x^2)sqrt(1-x^2)

f'(x) = -2/sqrt(1-x^2) +
2(1-2x^2)/(1-2x^2)sqrt(1-x^2)

f'(x) = -2/sqrt(1-x^2) +
2/sqrt(1-x^2)

f'(x) = 0

Since
the 1st derivative of the function is zero, therefore the function is a
constant.

For x = 0 => f(0) = arcsin 0 + 3arccos 0 +
arcsin 0

f(0) = 0 + 3*`pi` /2 +
0

f(0) = 3`pi`
/2

The given function f(x) is constant and
for x = 0 is f(0) = 3`pi` ` ` /2.