Question

Can somebody please check if my work is correct for the following question:

integral (2(xcosx-sinx))/x^2

=2 integral (xcosx-sinx)/x^2

=2 integral (xcosx)/x^2 - (sinx)/x^2

=2 integral cosx/x - 2 integral sinx/x^2

could somebody please explain it to me step by step?

Answer 1

@SolomonZelman

Answer 2

what is the next step if you could please show me

Answer 3

do you know how to take integral by part?

Answer 4

maybe i might remember if you showed me how

Answer 5

but please explain it to me step by step

Answer 6

for the first term, let u =1/x dv = cos x dx

Answer 7

ok, let me do it on paper, you can see it clearer

Answer 8

thanks

Answer 9

That's it

Answer 10

that is much better :)

Answer 11

but i have a question

Answer 12

You got it?

Answer 13

cant we let u = cosx?

Answer 14

yes i understood it thank you

Answer 15

as i said cant we also solve it the other way around by letting u=cosx?

Answer 16

You go nowhere on that way, because if you
let u = cos x , du = -sin xdx
dv = 1/x dx , v = lnx
so that the second term is
lnx cos x - \(\int \dfrac{lnx}{x}dx\)
and then??? have to use substitute again for this second term. Moreover, it is useless for \(\int \dfrac{sinx}{x^2}dx\)
You torture your life by going that way, hihihihihi

Answer 17

hehehehehe

Answer 18

but out of curiosity is it still right despite the torture lol

Answer 19

ok, show me how to take \(\int \dfrac{sin x}{x^2}dx\)

Answer 20

i would put it as integral 1/x^2 *sinx

Answer 21

then integral x^-2 * sinx

Answer 22

not to forget the dx at the end

Answer 23

hihihihi... I save your time by posting the answer from mathtool website:
http://www.wolframalpha.com/input/?i=int+%28sinx%2Fx^2%29+dx

Answer 24

what does Ci(x) stand for?

Answer 25

I don't know, it's out of our level. It may be for graduate level.

Answer 26

but even when i tried to integrate x^-2 * sinx i got x^-1/-1 * (-cosx)

Answer 27

which equals to cosx/x

Answer 28

i dont know where i got it wrong

Answer 29

nope, you let u = x^(-2) , right? then du = -2x^(-3) hehehe.... the integral becomes bigger and bigger.

Answer 30

ok so that would be my u

Answer 31

if i had to solve it

Answer 32

is there a trick to know which is the u?

Answer 33

ill let u know when i get back. thanks for your help :)

Answer 34

i now understand why thanks to you :) I really appreciate your help

Answer 35

can it be written like this:

2 integral cosx/x dx -2 integral sinx/x^2 dx

2 integral (sinx/x + integral sinx/x^2) -2 integral sinx/x^2 dx

Answer 36

@OOOPS

Answer 37

could ypu please help me @OOOPS