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How do you integrate y = e^x sin2x dx?

How do you integrate y = e^x sin2x dx?

Grade:12

1 Answers

grenade
2061 Points
8 years ago
Let: 

u = sin(2x) 
du = 2cos(2x) dx 
dv = e^x dx 
v = e^x 

Then: 

L = uv - ∫ v du 
==> L = e^x*sin(2x) - 2 ∫ e^x*cos(2x) dx 

Let: 

u = cos(2x) 
du = -2sin(2x) 
dv = e^x dx 
v = e^x 

By another round of integration by parts: 

L = e^x*sin(2x) - 2(uv - ∫ v du) 
==> L = e^x*sin(2x) - 2[e^x*cos(2x) + 2 ∫ e^x*sin(2x) dx] 
==> L = e^x*sin(2x) - 2e^x*cos(2x) - 4 ∫ e^x*sin(2x) dx 
==> L = e^x*sin(2x) - 2e^x*cos(2x) - 4L (since L = ∫ e^x*sin(2x) dx) 
==> 5L = e^x*sin(2x) - 2e^x*cos(2x) 
==> 5L = e^x * [sin(2x) - 2cos(2x)] 
==> L = e^x * [sin(2x) - 2cos(2x)]/5 
==> ∫ e^x*sin(2x) dx = e^x * [sin(2x) - 2cos(2x)]/5 + C 

Therefore, ∫ e^x*sin(2x) dx = e^x * [sin(2x) - 2cos(2x)]/5 + C. 
 
 

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