# Integration of x.tan(x)dx

1
by guptal

2015-06-06T11:06:53+05:30

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It is not possible to find the integral of x tan x as a function using the usual calculus methods.  So  we use the Taylor series - Maclaurin series for expansion of tan x:

the derivatives of tan x     and their values at x = 0 are:

f'(x) = sec^2 x               =>    f'(0) = 1
f''(x)=  2 sec^2 x tan x                =>    f''(0) =  0
f''' (x) =  4 sec^2 x tan^2 x + 2 sec^4 x         =>  f''' (0) = 2
f^(4) (x) =  8 sec^2 x tan^3 x + 8 sec^4 x tan x   + 8 Sec^4 x tan x          =>   f^4  (0) = 0
f^5 (x) = 16 sec^2 x tan^4 x + 24 sec^4 x tan^2 x + 8 sec^4 x tan^2 x + 8 sec^6 x
+ 8 sec^6 x + 32 sec^4 x tan^2 x       =>     f^(5) (0) = 16
Those terms in the derivatives that will be non-zero at 0 are :
f^(6) (x) =  0

So it can be given as a infinite terms polynomial.  Perhaps it converges to  a numerical value for limits between 0 and  π/2.

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