Question

求∫exsinx2dx

求∫exsinx2dx．

Answer 1

由题意可知：
∫e^xsinx²dx=

1

2

∫ex(1?cos2x)dx＝

1

2

ex?

1

2

∫excos2xdx      
∫e^xcos2xdx
=e^xcos2x+2∫e^xsin2xdx                
=e^xcos2x+2e^xsin2x-4∫e^xcos2xdx
∴∫excos2xdx＝

1

5

ex(cos2x+2sin2x)+C            
代入得：
原式=

1

2

ex?

1

10

ex(cos2x+2sin2x)+C