Question

1. - / 1.42 points evaluate the indefinite integral as an infinite series.

$$ int \frac { cos ( x ) - 1 } { x } d x $$
$$ sum _ { n = 1 } ^ { infty } ( square ) + c $$
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Explanation:

Step1: Recall the Maclaurin series of $\cos(x)$

The Maclaurin series of $\cos(x)=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}x^{2n}}{(2n)!}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots$
Then $\cos(x)-1=\sum_{n = 1}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}=-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots$

Step2: Divide by $x$

$\frac{\cos(x)-1}{x}=\sum_{n = 1}^{\infty}\frac{(-1)^{n}x^{2n - 1}}{(2n)!}=-\frac{x}{2!}+\frac{x^{3}}{4!}-\frac{x^{5}}{6!}+\cdots$

Step3: Integrate term - by - term

$\int\frac{\cos(x)-1}{x}dx=\int\sum_{n = 1}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n)!}dx$
Using the power - rule for integration $\int x^{m}dx=\frac{x^{m + 1}}{m+1}+C(m
eq - 1)$
$\int\frac{(-1)^{n}x^{2n-1}}{(2n)!}dx=\frac{(-1)^{n}x^{2n}}{(2n)!(2n)}+C$

Answer:

$\sum_{n = 1}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!(2n)}+C$