How do you find the antiderivative of #e^( -x)*cos2x#?
2 Answers
Explanation:
# int \ e^(-x) cos(2x) \ dx = e^(-x)/5{2sin(2x)-cos(2x)} + C#
Explanation:
Let:
# I = int \ e^(-x) \ cos(2x) \ dx #
We can use integration by parts:
Let
# { (u,=cos2x, => (du)/dx=-2sinx), ((dv)/dx,=e^(-x), => v=-e^(-x) ) :}#
Then plugging into the IBP formula:
# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #
gives us
# int \ (cos2x)(e^(-x)) \ dx = (cos2x)(-e^(-x)) - int \ (-e^(-x))(-2sin2x) \ dx #
# :. I = -e^(-x) cos2x - 2 \ int \ e^(-x) \ sin2x \ dx # .... [A]
At first it appears as if we have made no progress, as now the second integral is similar to
Let
# { (u,=sin2x, => (du)/dx=2cos2x), ((dv)/dx,=e^(-x), => v=-e^(-x) ) :}#
Then plugging into the IBP formula, gives us:
# int \ (sin2x)(e^(-x)) \ dx = (sin2x)(-e^(-x)) - int \ (-e^(-x))(2cos2x) \ dx #
# :. int \ e^(-x) \ sin2x \ dx = -e^xsin2x + 2I #
Inserting this result into [A] we get:
# I = -e^(-x) cos2x - 2 (-e^xsin2x + 2I) + A#
Now we can solve fior
# I = -e^(-x) cos2x + 2e^xsin2x - 4I + A#
# :. 5I = e^x{2sin2x - cos2x} + A#
# :. I = e^x/5{2sin2x - cos2x} + A#