Final answer:
The indefinite integral of e^(-t)sin(t)i + e^(-t)cos(t)j dt is -e^(-t)cos(t)i + e^(-t)sin(t)j + C + D.
Step-by-step explanation:
To find the indefinite integral of e(-t)sin(t)i + e(-t)cos(t)j dt, we can integrate each component separately.
The integral of e(-t)sin(t) with respect to t is -e(-t)cos(t) + e(-t)sin(t) + C, where C is the constant of integration.
The integral of e(-t)cos(t) with respect to t is -e(-t)sin(t) - e(-t)cos(t) + D, where D is the constant of integration.
Therefore, the indefinite integral of the given expression is -e(-t)cos(t)i + e(-t)sin(t)j + C + D.