MATH 2411 #7: QUESTION 7

#7: 7a)

1. ANSWER: du = 2e ^2x dx, dv = sinx dx and v = - cosx
   SOLUTION: e ^2x sinx dx = ( e ^2x )( sinx dx ) = u dv. Therefore u = e ^2x and dv = sinx dx. So du = 2e ^2x dx and v = - cosx
2. ANSWER:
   SOLUTION: Substituting all the parts into the formula results in
3. ANSWER: du = 2e ^2x dx, dv = cosx dx and v = sinx
   SOLUTION: e ^2x cosx dx = ( e ^2x )( cosx dx ) = u dv. Therefore u = e ^2x and dv = cosx dx. So du = 2e ^2x dx and v = sinx
4. ANSWER:
   SOLUTION:
5. ANSWER: (-1/5) e^2x cosx + (2/5) e^2x sinx
   SOLUTION: Let A be the integral. Then
6. ANSWER: (-1/5) e^2x cosx + (2/5) e^2x sinx + C