If you meant #inte^xsin^2(x)dx#, then this can be integrated.
We have to use the identity #sin^2(x)=1/2(1-cos(2x))#. The integral is then equal to:
#I=inte^xsin^2(x)dx=inte^x(1/2(1-cos(2x)))dx#
#color(white)I=1/2inte^xdx-1/2inte^xcos(2x)dx#
The first integral is elementary. We'll call the second one #J# and find it by itself:
#I=1/2e^x-1/2J#
For #J=inte^xcos(2x)dx#, we should use integration by parts. Let:
#{(u=cos(2x),=>,du=-2sin(2x)dx),(dv=e^xdx,=>,v=e^x):}#
So:
#J=e^xcos(2x)+int2e^xsin(2x)dx#
Perform integration by parts again.
#{(u=2sin(2x),=>,du=4cos(2x)dx),(dv=e^xdx,=>,v=e^x):}#
Then:
#J=e^xcos(2x)+2e^xsin(2x)-4inte^xcos(2x)dx#
Notice that #J# has reappeared on the right-hand side. We can add #4J# to both sides of the equation and then solve for #J#:
#5J=e^xcos(2x)+2e^xsin(2x)#
#J=1/5e^xcos(2x)+2/5e^xsin(2x)#
Returning to our original expression:
#I=1/2e^x-1/2J#
#color(white)I=1/2e^x-1/2(1/5e^xcos(2x)+2/5e^xsin(2x))#
#color(white)I=1/2e^x-1/10e^xcos(2x)-1/5e^xsin(2x)+C#
