Find the integral I of e^(2x)*cos*(x), with respect to x
Because we have a product of two functions of x, our first instinct is to apply integration by parts. Let u = e^(2x) and v' = cos(x). We then integrate v' to find v = sin(x) and differentiate u to find u' = 2e^(2x). Applying the by parts rule I = uv - (the integral of)(vu') we get I = e^(2x)sin(x) - 2(the integral of)(e^(2x)*sin(x)). The integral on the RHS is similar to the one we started with, so apply integration by parts again, this time with u = e^(2x), v' = sin(x), u' = 2e^(2x), v = -cos(x). This gives us I = e^(2x)sin(x) + 2e^(2x) - 4(the integral of)(e^(2x)cos(x)). The integral on the RHS is what we started with, so we substitute I in for it, getting I = e^(2x)sin(x) + 2e^(2x) - 4I. Rearranging and solving for I gives us I = e^(2x)(1/5)(sin(x) + 2cos(x)).
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