Log in to add a comment

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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.

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.