求sin(1+x^2)的泰勒展开式，包括带入的详细全过程

let
f(x)= sin(1+x) =>f(0) = sin1
f'(x)= cos(1+x) =>f'(0)/1! = cos1/1!
f''(x)=-sin(1+x) =>f''(0)/2! = -sin1/2!
f'''(x)=-cos(1+x) =>f'''(0)/3! = -cos1/3!
f^(4)(x)=sin(1+x) = f(x) =>f^(4)(0)/4! = sin1/4!
=>
sin(1+x)
=sin1- (cos1/1!)x - (sin1/2!)x^2 - (cos1/3!)x^3 + (sin1/4!)x^4 +.....
x =x^2
sin(1+x^2)
=sin1- (cos1/1!)x^2 - (sin1/2!)x^4 - (cos1/3!)x^6 + (sin1/4!)x^8 +.....

sin(1+x^2) = sin1cos(x^2) + cos1sin(x^2)
= sin1(1 - x^4/2! + x^8/4! - x^12/6! + ...) + cos1(x^2 - x^6/3! + x^10/5! - ...)