In position space,
             ⌠ ∞
    ❬φₙ❙Ψ❭ ≐ ⎮   dx φₙ⃰(x) Ψ(x).
             ⌡-∞

But for just 3L/4 to L, 
    ⌠L
    ⎮   dx φₙ⃰(x) Ψ(x).
    ⌡3L/4

Since this is a particle in an infinite square well,

    φₙ(x) = √⎛2͟⎞ sin⎛n͟π͟x͟⎞ and Ψ(x) = ∑ cₙ φₙ(x).
             ⎝L⎠    ⎝ L ⎠            ⁿ

So the probability to calculate is

    ⌠L
    ⎮   dx √⎛2͟⎞ sin⎛n͟π͟x͟⎞ ∑ cₙ φₙ(x).
    ⌡3L/4   ⎝L⎠    ⎝ L ⎠ ⁿ







The evaluation of the last piece of this expression is elucidated by the identity

    sin(u)sin(v) = 1/2 (cos(u-v) - cos(u+v)), i.e.

    sin(2nπ) sin(³/₂nπ) = ¹/₂(cos(1/2nπ) - cos(7/2nπ)).

So when n is odd, this product disappears, and when n is even one of the sin² terms disappears. The other sin² will always return 0, since regardless of the integer n, its argument is an integer multiple of π.