maclaurin series integral

Question

Answer 1

First use the Maclaurin series for sin(x)

sin(x) ≈ ∑(-1)^kx^2k+1/(2k+1)

Divide by x for sin(x)/x

sin(x)/x ≈ ∑(-1)^kx^2k/(2k+1)!

Integrate with respect to x --> ∫x^2kdx = x^2k+1/(2k+1) + C

Combine

∫sin(x)/x dx ≈ ∑(-1)^kx^2k+1/((2k+1)(2k+1)!) + C