What is the integral of (tanx)*(e^x) wrt x.?

Question

I cannot do it and all those I know believe it cannot be done in terms of known functions. Can you prove us wrong?

Angel heart: good try; except ... differentiating 1/2tanxe^x doesn't give tanx e^x. The mistake occurs in the 2nd integration by parts. The integral of
ln(secx) isn't tanx! (integral of sec^2x is tanx!!)

Answer 1

No, it can't be done in terms of elementary functions.
I'll try to concoct a proof for you, which I don't
have right now, but the main tool used in proving
this is a theorem of Liouville:
If f is an elementary function and
I = ∫ e^x f(x) dx is elementary
then
I must have the form Re^x,
where R is a rational function of x and tan x.
Let's try to apply this theorem here
Suppose I = ∫ e^x tan x is elementary
then
I = Re^x
Differentiate both sides to get
e^x tan x = e^x(R+R')
So R + R' = tan x.
Here's where I have to do some work to finish
the job.
How can one prove that there is no rational
function of x and tan x such that
R + R' = tan x?
For a way of doing this you might consut
the excellent article by Matthew Wiener at
http://www.math.niu.edu/~rusin/known-math/93_back/elementary.int

Answer 2

OOOPS MY BADD
getting lost in this notation...

*******************************************************************
OKAY this would be Easiest to do if I Scanned my hand calculations so as not to distract with all the--well you know. But here goes.
This problem is a bit of a Buggar !!!
Obviously you need to DO Integration by Parts. What is Most Annoying is one's Choice for u and for dv.
First of all Integration by Parts is given by:
Integral of ( u dv) = uv - Integral (v du )
Now obviously it would be great to begin by saying let
u = tanx and dv = e ^ x
so that
du = sec ^2x dx and v = e ^ x
Except you Soon discover Exactly where THAT goes to Naught !!!

SO begin with
u = e ^ x and dv = tanx dx

du = e ^x dx and v = ln (secx)

OH I know horrors that !!!
now you get
ln (secx) e ^ x - [ Integral (ln (secx) e ^x dx ]

NOW do the Integration by parts AGAIN for
[ Integral ( ln (secx) e ^x dx ]

u = e ^x dv = ln (secx) dx
du = e ^ x v = tanx

and again so uv - Integral v du becomes:

tanx e^x - Integral [ tanx e^x dx ]

NOW combine ALL :
Integral [ tanx e^x dx ] =
ln(secx) e^x - [ tanx e ^x - Integral [ tanx e^x dx]
carrying through the negative sign gives:
Integral [ tanx e^x dx ] =
ln(secx) e^x - tanx e ^x + Integral [ tanx e^x dx]
AND SO:

tanx e^x = ln(secx) e ^x
Integrate both sides of Equation

Integral [ tanx e ^x dx] = Integral [ ln(secx) e^x dx }
which is:
since we have already shown by Integration by Parts the
Integral [ ln(secx) e^x dx ] = tanx e^x - Integral[ tanx e^x dx ]

Integral [ tanx e^x dx] =
tanx e^x - Integral [ tanx e ^x dx ]
and SO

2 Integral [ tanx e^x dx ] = tanx e^x


Integral [tanx e^x dx ] = 1/2 [ tanx e^x} QED

WELL HOPING this above notation not TOO distracting...

I will attempt later to SCAN my hand calculations into document and cut and paste into my answer Next...

Answer 3

I agree this cannot be done in terms of known functions.

Answer 4

it cant be done in terms of elementary functions no. mathcad returns 'no symbolic result found' and mathematica returns something very complicated in terms of higher functions.

Answer 5

Use integration by parts, then the reduction formulae.

Answer 6

No I can't. I haven't a clue. But congratulations on asking your first question, and I hope you get lots of answers better than mine. Later Harry. :o)