to help calculate integral from 0 to t Sin[x]^2 dxThis question seems really easy but im having some difficulty with it.

This is what im thinking :

first put the [1-Cos[2x]]/2 instead of the Sin[x]^2

so this is what i have now (int from 0 to t) [1-Cos[2x]]/2dx

then i thought let u = cos[2x] then du = - Sin[2x]dx

and when x= 0 then u = 1 and when x= t then u= cos[2t]

then (int from 0 to t) [1-Cos[2x]]/2dx

=-1/2 (int from 1 to cos[2t] ) [1-u]/[Sin[2x]du]

Now i dont know what to do please help!!! :surprised