Help with integration?

Question

How did (e^x)(cos x)dx become 1/2[e^x(cos x + sin x)]? What method was used? I used integration by parts and couldn't get it.

Answer 1

Integration by parts should work:

Basic formula:

int(u dv) = uv - int(v du)

define:

u = e^x --> du = e^x dx

dv = cos(x) dx --> v = sin(x)

Applying the formula:

int( e^x cos(x) dx )

= e^x sin(x) - int( e^x sin(x) dx)

Apply Integration by parts one more time:

define:

u = e^x --> du = e^x dx

dv = sin(x) dx --> v = -cos(x)

= e^x sin(x) - [ -e^x cos(x) - int( -cos(x) e^x dx) ]

=e^x sin(x) + e^x cos(x) - int(e^x cos(x) dx)

Notice that the same integral is on both sides of the equation.  Just move the one on the right over to the left:

2 int(e^x cos(x) dx) = e^x (sin(x) + cos(x))

Therefore,

int( e^x cos(x) dx) = 1/2 e^x (sin(x) + cos(x))

Answer 2

integrate by parts:

∫u dv = uv - ∫v du

let u = e^x

du = e^x dx

let dv = cosx

v = sinx

∫e^x cosx dx = e^x sinx - ∫e^x sinx dx

integrate by parts again

let u = e^x

du = e^x dx

let dv = sinx

v = -cosx

∫e^x cosx dx = e^x sinx - [-e^x cosx - ∫-e^x cosx dx]

∫e^x cosx dx = e^x sinx + e^x cosx - ∫e^x cosx dx

add ∫e^x cosx dx for both sides

2 ∫e^x cosx dx = e^x sinx + e^x cosx

divide 2 for both sides

∫e^x cosx dx = (1/2) (e^x sinx + e^x cosx)

∫e^x cosx dx = (1/2)e^x (cosx + sinx)

Answer 3

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