Ackermann
Dec 27, 2020

answers: 1

Register to add an answer

The time for answering the question is over

Answer:

There is a pattern. Differentiating a function of the form eaxsin(bx) yields a linear combination of a function of the same form, and a function eaxcos(bx). The analogous property holds for functions eaxcos(bx). So the primitive of eaxsin(bx) will be a linear combination of eaxsin(bx) and eaxcos(bx) (plus a constant).

It remains to find the coefficients.

ddx(eax(msin(bx)+ncos(bx))=eax(a(msin(bx)+ncos(bx))+(bmcos(bx)−bnsin(bx)))=eax((am−bn)sin(bx)+(an+bm)cos(bx))

Now solve the linear system

am−bnan+bm=1=0.

Notice that

sin(bx)=Im(eibx)

so we need just take the imaginary part of this antiderivative

∫eaxeibxdx=1a+ibe(a+ib)x+C

Hint Integration by parts

∫u dv=uv−∫v du

Make substition

u=sin(bx) ⇒ du=bcos(bx) dx

and

dv=eax dx⇒v=eaxa

So

∫eaxsin(bx) dx=eaxasin(bx)−ba∫eaxcosbx dx

Then another integration by parts for ∫eaxcosbx dx . I think you can do the rest of it.

335

Chen

Dec 27, 2020

For answers need to register.

Contacts

mail@expertinstudy.com

Feedback

Expert in study

About us

For new users

For new experts

Terms and Conditions