Posted by kchik on Tuesday, December 4, 2012 at 12:09am.
if we let A be our integral, we have
A = ∫e^5t sin3t dt
u=e^5t
du = 5e^5t dt
dv = sin3t dt
v = -1/3 cos3t
A = uv - ∫v du
= -1/3 e^5t cos3t + 5/3 ∫e^5t cos3t dt
u=e^5t
du = 5e^5t dt
dv = cos3t dt
v = 1/3 sin3t
A = -1/3 e^5t cos3t + 5/3(1/3 e^5t sin3t - 5/3 ∫e^5t sin3t dt)
= -1/3 e^5t cos3t + 5/9 e^5t sin3t - 25/9 A
34/9 A = 1/9 e^5t (-3cos3t + 5sin3t)
A = 1/34 e^5t (-3cos3t + 5sin3t)
-----------------------------------
A = ∫xe^3x dx
u = x
du = dx
dv = e^3x dx
v = 1/3 e^3x
A = uv - ∫v du
= 1/3 xe^3x - 1/3 ∫e^3x dx
= 1/3 xe^3x - 1/9 e^3x
= 1/9 e^3x (3x-1)
------------------------
A = ∫x sec^2 x dx
u = x
du = dx
dv = sec^2 x dx
v = tan x
A = uv - ∫v du
= x tanx - ∫tanx dx
= x tanx - ∫sinx/cosx dx
= x tanx + log cosx
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