Question:

int_(sinx)^1(t^2.f(t))dt = (1-sinx) for x ∈ (0,pi/2)
Then f(1/sqrt(3)) = ?
a)3
b)sqrt(3)
c)1/3
d)"None"

Solution-1:
If we try to compare the LHS and RHS of the above equation, we see that if the function
(t^2.f(t)) = 1 , then it becomes; int_(sinx)^1(1.dt) = (1-sinx)

And that is equal to R.H.S.
So, we can say that here,
f(t) = 1/(t^2)

Therefore
f(1/sqrt(3)) = 3

Solution-2:
Upon differentiating the function,we get
(1.f(1).0 + sin^2(x).f(sinx).cosx) = (-cosx)

That is,
f(sinx) = 1/sin^2(x) Or f(p) = 1/p^2 where p = sin^2(x) for x ∈ (0,pi/2)