How do you integrate y = e^x sin2x dx?
Gman Namg , 9 Years ago
Grade 12
Integral Calculus> How do you integrate y = e^x sin2x dx?...
1 Answers
grenade
Last Activity: 9 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.
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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