Evaluate the indefinite integral as an infinite series.
[cos(x)-1]/x dx
Hey, get used to this
cos u --> 1 - u^2/2! + .....
routine
and the e^u --> 1
as u---> 0
Both of your problems are the same.
To evaluate the indefinite integral of [cos(x) - 1]/x dx as an infinite series, we can use a Taylor series expansion.
First, note that [cos(x) - 1]/x has a removable singularity at x = 0. This means that if we define the function as f(x) = (cos(x) - 1)/x for x ≠ 0 and f(0) = 0, then the integral of f(x) over any closed interval not containing 0 would be 0.
Now, let's focus on finding a Taylor series expansion for f(x) around x = 0. The Taylor series expansion for cos(x) centered at x = 0 is:
cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...
Substituting this into f(x), we get:
f(x) = [(1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...) - 1]/x
= -[(x^2)/2! - (x^4)/4! + (x^6)/6! - ...]/x
= -[x/2! - (x^3)/4! + (x^5)/6! - ...]
Now, we can integrate term by term in the series expansion of f(x) to obtain:
∫ f(x) dx = -∫ [x/2! - (x^3)/4! + (x^5)/6! - ...] dx
= -[(x^2)/2(2!) - (x^4)/4(4!) + (x^6)/6(6!) - ...] + C
This is the indefinite integral of [cos(x) - 1]/x dx expressed as an infinite series expansion. Note that C represents the constant of integration.