Integration prob ~ help plz

Don't integrals require a lowe and upper boundary?

Indefinite integrals don't

2 standard integrals are:

Integrate cos ^ 2 (x) dx = 1/2x + 1/4sin2x + Constant

and

Integrate cos ^ n (x) dx = 1/n cos ^(n-1) x sin x + (n-1)/n * Integrate cos ^ (n-2) x dx

So combine these chain itegrations to get

Integrate cos ^ 4(x)dx = 1/4 cos ^(3) x sin x + 3/4 [1/2x + 1/4sin2x + Constant]