about to only mathematics

Answer

3

Explanation

`F(x)=int_(x)^(x^(2)+(pi)/6)2cos^(2)tdt`
`:.F'(x)=2(cos(x^(2)+(pi)/6))^(2)2x-2cos^(2)x`
According to the question
`:.F'(a)+2int_(0)^(a)f(x)dx`
`implies2(cos(a^(2)+(pi)/a))^(2)2a-2cos^(2)a+2=int_(0)^(a)f(x)dx`
Defferentiating w.r.t `a` we get
`4cos^(2)(a^(2)+(pi)/6)+4axx2(a^(2)(pi)/6)(-sin(a^(2)+(pi)/6))`
`xx2a+4cos a sin a=f(a)`
`:. f(0)=4((sqrt(3))/2)^(2)=3`