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#.
