I FOUND e^(7x)[(7sin(4x)-4cos(4x))/65]+C
it is more simple ..check it
Hi, I thought I had this question until I got to the very last part, I believe this question uses the repeated integration by parts method, is that right?
Evaluate the indefinite integral:
u = e^7x
u' = (1/7)e^7x
v = (-1/4)cos(4x)
v' = sin(4x)
∫e^7x*sin(4x) = (-1/4)e^7x*cos(4x) - ∫(-1/4)cos(4x)*(1/7)e^7x
.....................= (-1/4)e^7x*cos(4x) + (1/28) ∫cos(4x)*e^7x <-- integrate this indefinite integral using integration by parts?
u = cos(4x)
u' = (-1/4)sin(4x)
v = (1/7)e^7x
v' = e^7x
And then I get...
(1/28) ∫cos(4x)*e^7x = (1/28)(cos(4x)*(1/7)e^7x + ∫(1/7)e^7x*(-1/4)sin(4x))
..............................= (1/196)(cos(4x) + (1/28^2)∫e^7x*sin(4x) <-- so this second term is now the same as the original equation, except for the constant of course, and I should just simplify, right? But I don't know the fractions make it look really complicated and I don't think I'm doing it right...
Does someone know where I went wrong and how to solve it properly?
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