integration by parts?

Question

prove:-
(1)int(e^(ax)sin(bx))dx
=e^(ax){(asinbx-bcosbx)/
(a^2+b^2)}+C
(2)int(e^(ax)cos(bx))dx
=e^(ax){(acosbx+bsinbx)/
(a^2+b^2)}+C

-a similar question was asked
yesterday,however,this is
more difficult

-use integration by parts from
first principles and you will
get it!

ra

there's a lot of algebra
in this-a lot of pitfalls,but
take your time and you
will get the right answer
thanks

Answer 1

A way to do both of these integrals at the same time is to realize that they are the imaginary and real components to the integral of
e^[(a+bi)x]=e^(ax)cos(bx)+i*e^(ax)sin(bx).
This integral is, of course,
1/(a+bi) e^[(a+bi)x] +C.

If you now rationalize the denominator and break into real and imaginary parts, you get what you want.

Answer 2

1)

Integral (e^(ax) sin(bx) ) dx

[Note: This is as hard as integrals get!!]. By the way, since I know what's going to happen with this integral, I'm going to call this integral "S".

So we solve it by parts.

Let u = e^(ax). dv = sin(bx) dx
du = a e^(ax). v = (-1/b) cos(bx)

S = e^(ax) (-1/b) cos(bx) - Integral (a e^(ax) (-1/b) cos(bx) )dx

Let's clean it up a bit by moving all constants to the front. Note that we can safely pull out constants OUTSIDE of the integrla as well.

S = (-1/b) e^(ax) cos(bx) - (-a/b) Integral (e^(ax) cos(bx) )dx

Two minuses make a plus.

S = (-1/b) e^(ax) cos(bx) + (a/b) Integral (e^(ax) cos(bx) )dx

Now, we solve the second integral by parts.

Let u = e^(ax). dv = cos(bx) dx
du = a e^(ax). v = (1/b)sin(bx)

S = (-1/b) e^(ax) cos(bx) + (a/b) [e^(ax) (1/b) sin(bx) -
Integral ( (a/b) e^(ax) sin(bx) ) dx ]

Distributing the (a/b) over the brackets,

S = (-1/b) e^(ax) cos(bx) + (a/(b^2)) e^(ax) sin(bx) -
([a^2]/[b^2]) Integral (e^(ax)sin(bx)) dx

Apparently, it appears that we appeared right back where we started. However, this is not a problem at all, because
Integral (e^(ax) sin(bx))dx = S! All we have to do is replace it by S.

S = (-1/b) e^(ax) cos(bx) + (a/(b^2)) e^(ax) sin(bx) - ([a^2]/[b^2]) S

Now, we can move that term with the S on the right hand side, to the left hand side,

S + ([a^2]/[b^2])S = (-1/b) e^(ax) cos(bx) + (a/(b^2)) e^(ax) sin(bx)

Factoring out an S,

S(1 + (a^2)/(b^2)) = (-1/b) e^(ax) cos(bx) + (a/(b^2)) e^(ax) sin(bx)

Simplfying what's in the brackets,

S((a^2 + b^2)/(b^2)) = (-1/b) e^(ax) cos(bx) + (a/(b^2)) e^(ax) sin(bx)

And now, to isolate S, we have to multiply both sides by
(b^2)/(b^2 + a^2). Since this is a constant, all we have to do is multiply this by each constant coefficient. You'll notice that (-1/b) will cancel with part of (b^2 / (b^2 + a^2) ), and so forth.

S = [-(a^2 + b^2)/b] e^(ax) cos(bx) + [a/(b^2 + a^2)] e^(ax)sin(bx)

And this is our answer, because remember what we let S equal!

S = Integral (e^(ax) sin(bx) ) dx

So we WANTED to solve for S, and there it is!