Question
Fri November 02, 2012

# Integrate e^x(sinx-cosx)

Sat November 03, 2012
Using the trigonometric identity
sin(2x) = 2sin(x)cos(x)
giving
sin(x)cos(x) = (1/2)sin(2x)

we have
intg(ex)sin(x)cos(x) dx = (1/2) intg(ex)sin(2x) dx

we can now integrate by parts.

Let
u = sin(2x)
then
du/dx = 2cos(2x)
and
dv/dx = ex
so
v = ex

Hence
(1/2) intg (ex) sin(2x) dx
= (1/2)[ (ex)sin(2x) - 2 intg (ex)cos(2x) dx]
= (1/2)(ex)sin(2x) - intg (ex)cos(2x) dx ............... (a)

then integrate this integral by parts

Let u = cos(2x)
then
du/dx = -2sin(2x)
and
dv/dx = ex = v

So
intg (ex)cos(2x) dx
= (ex)cos(2x) - 2 intg (ex)(-sin(2x)) dx
= (ex)cos(2x) + 2 intg (ex)sin(2x) dx

Substitute this into the original equation (a)

(1/2)intg(ex)sin(2x) dx = (1/2)(ex)sin(2x) - (ex)cos(2x) - 2intg(ex)sin(2x) dx

Notice that on both sides of the equation we have a intg(ex)sin(2x) dx term. Take both of these terms to the same side to get:

(1/2)intg(ex)sin(2x) dx + 2intg(ex)sin(2x) dx= (1/2)(ex)sin(2x) - (ex)cos(2x)

(5/2) intg (ex)sin(2x) dx = (1/2)(ex)sin(2x) - (ex)cos(2x)

divide both sides by 5

(1/2) intg (ex)sin(2x) dx = (1/5)(ex)[(1/2)sin(2x) - cos(2x)]

therefore
intg(ex)sin(x)cos(x) dx = (1/5)(ex)[(1/2)sin(2x) - cos(2x)] +C Answer
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