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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


f(x) = f(0) + f'(0)+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+....+\frac{f^{(n)}(0)}{n!}x^n+....\\\\tan\ x=x+\frac{x^3}{3}+\frac{2}{15}x^5+\frac{17}{315}x^7+\frac{62}{2835}x^9+....\\\\ \int\limits^{}_{} {x\ tan\ x} \, dx =\frac{x^3}{3}+\frac{x^5}{15}+\frac{2}{105}x^7+\frac{17}{2835}x^9+\frac{62}{31185}x^{11}....

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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