How do you evaluate the integral #int e^-xcosx#?

1 Answer
Jun 6, 2017

Answer:

# int \ e^(-x) \cosx \ dx = 1/2(e^(-x)sinx - e^(-x)cosx) + C #

Explanation:

Let:

# I = int \ e^(-x) \cosx \ dx #

We can use integration by parts:

Let # { (u,=cosx, => (du)/dx,=-sinx), ((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 \ (cosx)(e^(-x)) \ dx = (cosx)(-e^(-x)) - int \ (-e^(-x))(-sinx) \ dx #
# :. I = -e^(-x)cosx - int \ e^(-x) \ sinx \ dx # .... [A]

At first it appears as if we have made no progress, as now the second integral is similar to #I#, having exchanged #cosx# for #sinx#, but if we apply IBP a second time then the progress will become clear:

Let # { (u,=sinx, => (du)/dx,=cosx), ((dv)/dx,=e^(-x), => v,=-e^(-x) ) :}#

Then plugging into the IBP formula, gives us:

# int \ (sinx)(e^(-x)) \ dx = (sinx)(-e^(-x)) - int \ (-e^(-x))(cosx) \ dx #
# :. int \ e^(-x) \ sinx \ dx = -e^(-x)sinx + I #

Inserting this result into [A] we get:

# I = -e^(-x)cosx + e^(-x)sinx - I + A #

# :. 2I = e^(-x)sinx - e^(-x)cosx + A #
# :. \ \ I = 1/2(e^(-x)sinx - e^(-x)cosx) + C #