Integrate[ (Cos[Pi x /L])^4 (2/L)^2, {x, -L/2, L/2}]

FullSimplify[Integrate[Cos[(Pi*x)/L]^2*(2/L)^2*Cos[((2*n + 1)*Pi*x)/L]*
    Cos[((2*m + 1)*Pi*x)/L], {x, -(L/2), L/2}], Element[{m, n}, Integers]]

FullSimplify[Integrate[ (Cos[Pi x /L])^2 (2/L)^2 Cos[(2 n +1)Pi x/L] Sin[(2 m )Pi x/L], {x, -L/2, L/2}], {m, n} ∈ Integers]

FullSimplify[Integrate[ (Cos[Pi x /L])^2 (2/L)^2 Sin[(2 n )Pi x/L] Sin[(2 m )Pi x/L], {x, -L/2, L/2}], {m, n} ∈ Integers]

FullSimplify[Integrate[ (Cos[Pi x /L])^2 (2/L)^2 Sin[(2 n )Pi x/L] Cos[(2 m +1 )Pi x/L], {x, -L/2, L/2}], {m, n} ∈ Integers]

Integrate[ (Cos[Pi x /L])^2 (Cos[Pi y /L])^2 (2/L)^2, {x, -L/2, L/2}, {y, -L/2, L/2}]

Integrate[(2/L)^2*Cos[(Pi*x)/L]^2*Cos[(3*Pi*x)/L]^2, 
  {x, -(L/2), L/2}]

Integrate[ (Cos[Pi x /L])^2 (2/L)^2 Cos[(2 n +1)Pi x/L] Cos[(2 m +1 )Pi x/L], {x, -L/2, L/2}]

(* Note the trouble spots where m -n = 0 or (+-)1 + m -n = 0 

asking about this here:

http://stackoverflow.com/questions/7743774/proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints

where the following L=2 simplification was used.
*)

Integrate[Cos[(Pi*x)/2]^2*Cos[(3*Pi*x)/2]^2, 
  {x, -1, 1}]

FullSimplify[Integrate[Cos[(Pi*x)/2]^2* Cos[((2*n + 1)*Pi*x)/2]*
    Cos[((2*m + 1)*Pi*x)/2], {x, -1, 1}], Element[{m, n}, Integers]]