∫xtanx dx

I'm trying but cant reach anywhere

votz da answer?is it xln/cosecx/ -∫ ln/cosx/?im not gettin the value of II

please let me know the answer

if we take x=first function
tanx=second function

then

let, I1= ∫xtanxdx= x∫tanxdx - ∫{1.∫tanxdx}dx
= xln|secx| - ∫ln|secx|dx
=xln|secx| - I2 ----------------------------(1)

I2 = ∫1.ln|secx|dx = ln|secx|∫dx - ∫{(secxtanx/secx)∫dx}dx
= xln|secx| - ∫xtanx dx
= xln|secx| - I1 ------------------------------(2)

from (1) and (2) we can see dat this function is retraced........ i mean ulta wahi ho jaara....
so i dun think this is possible....
from where did u get?? CBSE2008 by chance?? 'coz we got a question last time .... it was somewat like this....