急~問題 請求出a的值?

I have seen this episode on youtube too.
a can be any real number. Therefore I guess the question ask for the value of a when the area of f(x) is smallest.
I only have knowledge in high school to solve it. If you have better solution, please tell me.
f(x) = integrate | sin(x) - a cos(x)| dx from 0 to π/2.
Consider the x-intercept,
tanx = a
For 0 <= x-intercept < π/2 , a>0
For π/2 < x-intercept < π ,a<0

For a<0, π/2 < x-intercept <= π
f'(x) = cosx + asinx < 0
y- intercept = -a>0
Therefore, the graph is positive from 0 to π/2.

Area: integrate sinx - a cosx dx from 0 to π/2.
= 1 - a >= 1.

For a>0, 0<= x-intercept < π/2
f'(x) = cosx + asinx >0
y- intercept = -a<0
The graph is negative from 0 to x-intercept ( Arctan(a) ) , positive from x-intercept ( Arctan(a) ) to π/2.
The absolute are is then:
[integrate acosx - sinx dx from 0 to Arctan(a) ] + [integrate sinx - acosx dx from Arctan(a) to π/2.]

Area, f(a) = 2a sin(Arctan(a)) + 2cos(Arctan(a)) -1 -a
f'(a)=0 when the value of area is smallest/ largest.
sin(Arctan(a)) + acos(Arctan(a))(1/(a^2+1))-sin(Arctan(a))(1/(a^2+1)) = 1/2
Sub a = root3 /3, LHS=RHS, and f(root3/3) < 1 ,f''(root3/3) >0
Therefore a = root3 /3