Question

Evaluate the indefinite integral as an infinite series ∫sinx /2x dx. Find the first five non-zero terms of series representation centered at x=0. f(x)=

Answer 1

you will want to start with the series for sin(x) ....

Answer 2

i got the first one correct, i put 0

Answer 3

i think the second one should be \[\frac{ x ^{3} }{ 36 }\]

Answer 4

okay so the second appears to be \[\frac{ 1 }{ 2x }\]

Answer 5

\[\frac{ x }{ 2 }\] i mean

Answer 6

i see that the notifications are not working either :/

Answer 7

\[\sin x=\sum_0\frac{(-1)^n}{(2n+1)!} x^{2n+1}\]
\[\frac{\sin x}{2x}=\frac12\sum_0\frac{(-1)^n}{(2n+1)!} x^{2n}\]
\[\int \frac{\sin x}{2x}=\frac12\int~\sum_0\frac{(-1)^n}{(2n+1)!} x^{2n}\]
\[\int \frac{\sin x}{2x}=\frac12\sum_0~\int\frac{(-1)^n}{(2n+1)!} x^{2n}\]
\[\int \frac{\sin x}{2x}=\frac12\sum_0~\int\frac{(-1)^n}{(2n+1)!~(2n+1)} x^{2n+1}+C\]right?

Answer 8

pfft, forgot to cancel out the int sign :)

\[\int \frac{\sin x}{2x}=\frac12\sum_0\frac{(-1)^n}{(2n+1)!~(2n+1)} x^{2n+1}+C\]

Answer 9

\[\frac12\left(\frac{1}{1!~1}x^{1}-\frac{1}{3!~3}x^{3}+\frac{1}{5!~5}x^{5}-\frac{1}{7!~7}x^{7}\pm...\right)+C\]

Answer 10

yes thanks!! its correct :D i was missing the negative sign :P

Answer 11

youre welcome