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? 

Expert Answers info

sciencesolve eNotes educator | Certified Educator

calendarEducator since 2011

write5,349 answers

starTop subjects are Math, Science, and Business

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 =...

(The entire section contains 301 words.)

Unlock This Answer Now


check Approved by eNotes Editorial