How do you find the integral of e^(-x)sin(2x)dx using integration by parts? Here's my work: u = sin(2x) dv to be e^(-x)dx du = 2cos(2x)dx v = -e^(-x) integral of u(dv) = uv - integral of v(du) substituing values, i get integral of e^(-x)sin(2x)dx = -sin(2x)e^(-x) - integral of -2cos(2x)e^(-x) after that, trying to find the the integral of the last part just leaves me in a perpetual cycle of integration :( what am I doing wrong?
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Luca B.
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You need to avoid the perpetual cycle, hence, you should use the following notation `int ``e^(-x)sin 2x dx` = I and you need to use parts to solve the integral such that:
`u = e^(-x) => du = -e^(-x)dx`
`dv = sin 2x dx => v = -(cos 2x)/2`
Using the following formula yields:
`int udv = uv - int vdu`
`int e^(-x)sin 2x dx = -(e^(-x)*cos 2x)/2 - (1/2)int e^(-x)cos 2x dx `
You should solve the integral `int e^(-x)cos 2x dx` using parts such that:
`u =...
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