How can you solve this?

Question

Integral Xtanx dx

Answer 1

Integrate by parts!
Let f(x) = tanx and g'(x) = x
Then integral of f(x)g'(x) = f(x)g(x) - integral of f'(x)g(x)

or,
solution x^2tanx/2 - integral of f'(x) (but the integral of f'(x) is simply f(x))

solution is x^2tanx/2 - tanx

Answer 2

Integration by parts doesn't work because
∫x tan x dx
=(x^2/2)tan x - (1/2)∫x^2sec^2x dx

The second part is more complicated.

One possible approach for the problem is to expand tan x as a Taylor series at x = 0, and then take integral.

Answer 3

Sorry! This integral is not elementary.
If you integrate it by parts, you end up
having to integrate ln(cos x). The answer
to that involves the dilogarithm function.
Here is the answer to your original problem
from integrals.wolfram.com:
ix²/2 -x log(1 + e^(2ix) ) + i/2*Li_2(e ^(-2ix) ).
Here Li_2 is the dilogarithm function.
I'll come back here later and try to
find a form of the answer which doesn't
involve imaginary quantities.

Answer 4

if x=u
tanx=v' v'=dv/dx
then integ xtanx
=uv-integ(u'*v)