Notice that x+2a^2-a is just x with the added constant 2a^2-a. This makes the substitution u=x+2a^2-a not too difficult, since du=dx.
We can substitute this directly into the integral, but remember that the bounds will change as we move from x to u. The bound x=a will become u=(a)+2a^2-a=2a^2 and the bound x=-a becomes u=(-a)+2a^2-a=2a^2-2a.
We then see that:
int_(-a)^acos(x+2a^2-a)dx=int_(2a^2-2a)^(2a^2)cos(u)du
The integral of cosine is just sine:
=[sin(u)]_(2a^2-2a)^(2a^2)=color(blue)(sin(2a^2)-sin(2a^2-2a)
Which is not equivalent to what you stated. You may have attempted to do [sin(u)]_(2a^2-2a)^(2a^2)=sin(2a^2-(2a^2-2a)), but this is not how to properly evaluate the function.
You could also get a "simpler" answer using the sine difference to product formula: sin(alpha)-sin(beta)=2cos((alpha+beta)/2)sin((alpha-beta)/2).
Here that gives the reduction:
=2cos((2a^2+2a^2-2a)/2)sin((2a^2-(2a^2-2a))/2)=2cos(2a^2-a)sin(-a)
Upon re-examining this, perhaps you want a solution for a.
In that case, we want that 2cos(2a^2-a)sin(-a)=-sin(2a).
Note that -sin(2a)=-2cos(a)sin(a). Also note that 2cos(2a^2-a)sin(-a)=-2cos(2a^2-a)sin(a). So, the two are equal when cos(2a^2-a)=cos(a), which occurs when 2a^2-a=a.
Solving this yields 2a^2-2a=2a(a-1)=0, so a=0 or a=1.
