Hi, could someone please help me with this question? Thanks

Question

Calculate the Maclaurin series of f(x)=4x^3*sin(x^4). Explicitly use the Maclaurin series for sine. Using this series for f(x), verify that the integral of 4x^3*sin(x^4)dx=-cos(x^4)+C.

Answer 1

we know that

sin(u) = u - u^3/3! + u^5/5! - ...
so, sin(x^4) = x^4 - x^12/3! + x^20/5! - ...
so, f(x) = 4x^7 - 4x^15/3! + 4x^23/5! - ...

it is certainly obvious that d/dx -cos(x^4) = 4x^3 sin(x^4). Using series,

-cos(x^4) = -1 + x^8/2! - x^16/4! + x^24/6! + ...
so, its derivative is
8x^7/2! - 16x^15/4! + 24x^23/6! - ...
= 4x^7 - 4x^15/3! + 4x^23/5! - ...

and they are the same