Question

Topic: \(\textbf{Calculus 2}\)

Evaluate the indefinite integral as an infinite series.

\[\int \frac{\cos x − 1}{x} \ \ \ dx \]

What the heck? I mean putting these in Taylor series forms are hard enough, but this? Not even sure where to start here >_< It worries me because the final is coming up and this is a hole in my knowledge... So, um, help?

Answer 1

Well, could you start with the infinite series for cosine x?

Answer 2

You'd have (1/x) (-x^2/2! + x^4/4! .....)

Answer 3

\[ \int \frac{\cos x}{x}dx - \int \frac{1}{x}dx \]

Answer 4

Let's see... \[\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, \text{for all x = } \mathbb{R}\]

Answer 5

That's the MacLaurin equivalent for cosine

Answer 6

\[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \]

Answer 7

Is it? Did I look at the wrong one? (-:

Answer 8

oh yes that's the hyperbolic one! SOrry!

Answer 9

Oh no wait I have the wrong one, I wrote sine by mistake

Answer 10

Arg soo confusing lol

Answer 11

lets both check again then hehe.

Answer 12

yes I believe I have the right one

Answer 13

I thought you couldn't just do 1/x , you'd have to convert that to a series as well, yes?

Answer 14

\[ \cos x = \sum_{n = 0}^\infty {(-1)^nx^{2n}\over (2n)!}\]
which makes
\[ \int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = {(-1)^nx^{2n}\over 2n(2n)!}\]

Answer 15

+c

Answer 16

Anti-derivative power rule?

Answer 17

\[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \]
Distribute it out

\[ \frac{\cos x}{x} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n-1}}{(2n)!} \]

Answer 18

yes that's it @agentx5

Answer 19

all has mistake!

Answer 20

where @mahmit2012 ?

Answer 21

all !but it is time to pray i wll come back and show you

Answer 22

\[ \int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = \sum_{n=1}^\infty {(-1)^nx^{2n}\over 2n(2n)!} \]

Answer 23

after pray !

Answer 24

lol ... where did i make mistake??

Answer 25

Did we do the same @experimentX ? besides that you completed it already.

Answer 26

I think we did.

Answer 27

n=1, n=0

Answer 28

yes but obviously "ALL" is wrong :D

Answer 29

first of all the original question has interval [0,inf)
and it has many different solution to get the answer.
and if you are looking for undefined integral so just a matching series is the answer ! but with a log function.