Evaluate the indefinite integral as an infinite series ∫ cos x-1/x dx.

Solution:

Given function is f(x) = (cosx - 1) / x

By using Maclaurin's series,

the expansion of f(x)= cosx = \(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+----(-1^{n})\frac{x^{2n}}{(2n)!}\)

⇒The expansion of f(x)= cosx =\( \sum_{n=0}^{\infty}(-1^{n\ })\frac{x^{2n}}{(2n)!} \)

i.e cosx - 1 = \( \sum_{n=0}^{\infty}(-1^{n\ })\sum_{n=0}^{\infty}(-1^{n\ })\frac{x^{2n}}{(2n)!} - \frac{x^{0}}{(0)!}cosx - 1 = \sum_{n=1}^{\infty}(-1^{n\ })\frac{x^{2n}}{(2n)!} \) {on simplification}

i.e\( \frac{cosx - 1 }{x} = \frac{\sum_{n=1}^{\infty}(-1^{n\ })\frac{x^{2n}}{(2n)!}}{x} \)

i.e\( \frac{cosx - 1 }{x} = \sum_{n=1}^{\infty}(-1^{n\ })\frac{x^{2n-1}}{(2n)!} \)

Consider \( \int \frac{cosx - 1 }{x} = \int \sum_{n=1}^{\infty}(-1^{n\ })\frac{x^{2n-1}}{(2n)!} \)

We know that \( \int x^{2n-1} = \frac{x^{2n-1+1}}{(2n-1+1)}+c \)

Therefore, \( \int \frac{cosx - 1 }{x}=\sum_{n=1}^{\infty}(-1^{n\ })\frac{x^{2n}}{(2n)!(2n)}+c \)

Where c is the constant of integration.

Evaluate the indefinite integral as an infinite series ∫ cos x-1/x dx.

Summary:

The indefinite integral as an infinite series ∫ (cos x-1)/x dx is

\( \int \frac{cosx - 1 }{x} = \sum_{n=1}^{\infty}(-1^{n\ })\frac{x^{2n}}{(2n)!(2n)}+c. \)