Problem
Question 4 5 Points Use the Maclaurin series representation of $\sin (x)$ to find the Maclaurin series representation of $\int \frac{\sin (x)}{x^{2}} d x$. A $C+\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n+1)(2 n+2) !}$ (B) $C+\frac{1}{x^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n+2}}{(2 n+2) !}$ (C) $C+\ln (1 x))+\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n)(2 n+1) !}$ (D) $\sum_{n=1}^{\infty} \frac{(-1) n(2 n)}{(2 n)(2 n+1) !}$
Solution
Step 1 ：First, let's divide the Maclaurin series of sin(x) by $x^2$:
Step 2 ：\(\frac{\sin(x)}{x^2} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n-1}}{(2n+1)!}\)
Step 3 ：Now, we need to integrate this series term by term:
Step 4 ：\(\int \frac{\sin (x)}{x^{2}} d x = C + \sum_{n=0}^\infty \int \frac{(-1)^n x^{2n-1}}{(2n+1)!} dx\)
Step 5 ：After integration, the general term should be:
Step 6 ：\(\frac{(-1)^n x^{2n}}{(2n+1)(2n+2)!}\)
Step 7 ：\(\boxed{\text{Final Answer: } A}\)
