Question

Evaluate the indefinite integral as an infinite series. 45.\(\int {\frac{{cosx - 1}}{x}dx} \)

Short answer

\(\int {{x^2}\sin ({x^2})dx = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)(2n)!}} + C} } \)

Step-by-step solution

Step 1: Explanation

We evaluate the given indefinite integral as an expression of an infinite series. Here we are going to use Maclaurin series representation of some functions and work our way through to obtain similar pattern of expression. Given indefinite integral

\(\int {\frac{{\cos x - 1}}{x}dx} \)

Step 2: Recall and use Maclaurin series representation

Recall that the Maclaurin series representation for \(\cos x\) is given by

\(\cos (x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} \)

It follows that

\(\begin{array}{l}\cos (x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} \\\cos (x) - 1 = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} - 1\\\cos (x) - 1 = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} \end{array}\)\(\begin{array}{l}\frac{{\cos (x) - 1}}{x} = \sum\limits_{n = 1}^\infty {\left( {\frac{1}{x}}\right)\frac{{{{(1)}^n}{x^{2n}}}}{{(2n)!}}} \\\frac{{\cos (x) - 1}}{x} = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{2n - 1}}}}{{(2n)!}}} \end{array}\)

Step 3: Represent indefinite integral as infinite series

Therefore the indefinite integral is represented as an infinite series as shown below:\(\int {{x^2}\sin ({x^2})dx = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)(2n)!}} + C} } \)