How do you find the integral of e^(-x)sin(2x)dx using integration by parts?Here's my work: u = sin(2x) dv to be e^(-x)dx du = 2cos(2x)dx  v = -e^(-x) integral of u(dv) = uv - integral of v(du)...

How do you find the integral of e^(-x)sin(2x)dx using integration by parts?

Here's my work:
u = sin(2x)
dv to be e^(-x)dx
du = 2cos(2x)dx 

v = -e^(-x)

integral of u(dv) = uv - integral of v(du)

substituing values, i get

integral of e^(-x)sin(2x)dx = -sin(2x)e^(-x) - integral of -2cos(2x)e^(-x)

after that, trying to find the the integral of the last part just leaves me in a perpetual cycle of integration :( what am I doing wrong? 

Asked on by modesthief

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to avoid the perpetual cycle, hence, you should use the following notation `int ``e^(-x)sin 2x dx`  = I and you need to use parts to solve the integral such that:

`u = e^(-x) => du = -e^(-x)dx`

`dv = sin 2x dx => v = -(cos 2x)/2`

Using the following formula yields:

`int udv = uv - int vdu`

`int e^(-x)sin 2x dx = -(e^(-x)*cos 2x)/2 - (1/2)int e^(-x)cos 2x dx `

You should solve the integral `int e^(-x)cos 2x dx`  using parts such that:

`u = e^(-x) => du = -e^(-x)dx`

`dv = cos 2x dx=> v = (sin 2x)/2`

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

Substituting `(e^(-x)*sin 2x)/2 + (1/2)int e^(-x)(sin 2x)dx ` for `int e^(-x)cos 2x dx ` yields:

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

Using the notation `int e^(-x)(sin 2x)dx = I`  yields:

`I = -(e^(-x)*cos 2x)/2 - (e^(-x)*sin 2x)/4 - (1/4)I`

You should isolate the terms that contain I to the left side such that:

`I + I/4 =-(e^(-x)*cos 2x)/2 - (e^(-x)*sin 2x)/4`

`5I/4 =-(e^(-x)*cos 2x)/2 - (e^(-x)*sin 2x)/4`

`5I = -(2e^(-x)*cos 2x) - (e^(-x)*sin 2x)`

`I = (-(2e^(-x)*cos 2x) - (e^(-x)*sin 2x))/5`

Substituting back `int e^(-x)(sin 2x)dx`  for I yields:

`int e^(-x)(sin 2x)dx = (-(2e^(-x)*cos 2x) - (e^(-x)*sin 2x))/5`

Hence, evaluating the given integral using parts yields `int e^(-x)(sin 2x)dx = (-(2e^(-x)*cos 2x) - (e^(-x)*sin 2x))/5` .

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