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?
- print Print
- list Cite
Expert Answers
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
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- Find the indefinite integral `int x sqrt((2x-1)) dx` using integration by substitution `int x...
- 1 Educator Answer
- (e^2x+1)/e^x dxintegrate by substitution method
- 1 Educator Answer
- Evaluate the indefinite integral integrate of sin^4(x)cos^4(x)dx
- 1 Educator Answer
- Find the integral integrate of sin(5x)cos(2x) dx
- 1 Educator Answer
- Evaluate the integral integrate of (sin(x))^2(cos(x))^4 dx
- 1 Educator Answer