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 - integral sinx/x^2

=2 integral 1/x * cosx - x^-2 * sinx

=2log(x)sinx-1/x * cosx +c

Answer 1

This is your answer, the last line ?

Answer 2

yes

Answer 3

i believe it still can be simplified

Answer 4

but im not sure

Answer 5

I am not so sure, what you are doing after

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

Answer 6

2 integral 1/x * cosx - integral x^-2 * sinx

Answer 7

2 integral[ 1/x * cosx - x^-2 * sinx]
upto this step its alright
now from here u should use product rule of integration on integral[[ 1/x * cosx]
\[\int\limits_{}^{}uvdx=u \int\limits_{}^{}vdx-\int\limits_{}^{}[\frac{ d }{ dx }(u)*\int\limits_{}^{}vdx]dx\]

Answer 8

it is supposed to be
2 integral 1/x * cosx - 2 integral x^-2 * sinx

not just
2 integral 1/x * cosx - integral x^-2 * sinx

Answer 9

2(a-b) is not equal to 2a-b

Answer 10

yeah

Answer 11

\(\LARGE\color{blue}{ 2\int_{ }^{ } \frac{\cos(x)}{x}- 2\int_{ }^{ } \frac{\sin(x)}{x^2} }\)

Answer 12

then you get

\(\Large\color{blue}{ 2\int_{ }^{ } \frac{1}{x}\cos(x)- 2\int_{ }^{ } \frac{1}{x^2}\sin(x) }\)

Answer 13

ok now i see where i made the mistake. so in other words if i have a constant outside the integral that constant also applies to the second integral?

Answer 14

yeah now the first intrgral will be
\[(1/x)\int\limits_{}^{}cosxdx-\int\limits_{}^{}[(-1/x^2)(-sinxdx)\]
and look they are getting cancelled out

Answer 15

*it will be only (sinxdx) not (-sinxdx)

Answer 16

@sidsiddhartha you can not take 1/x out, it is not a constant

Answer 17

you did that in your last black latex integral.

Answer 18

i agree solomon x's are supposed to be in not out

Answer 19

I would just go with

\(\Large\color{blue}{ 2\int_{ }^{ } \frac{\cos(x)}{x}- 2\int_{ }^{ } \frac{\sin(x)}{x^2} }\)

Answer 20

so it should be 2logx sinx - 2 (x^-1/-1) * -cosx

Answer 21

@SolomonZelman i'm not taking i/x out
i'm using uv substitution in which 1/x is =u
and cosx=v

Answer 22

just see the formula i wrote above for uv substitution @SolomonZelman

Answer 23

2logxsinx-2/xcosx

Answer 24

would that be the answer?

Answer 25

\(\Large\color{blue}{ 2\int_{ }^{ } \frac{\cos(x)}{x}- 2\int_{ }^{ } \frac{\sin(x)}{x^2} }\)

\(\Large\color{blue}{ [2Ci(x)+C]~~- 2(~~Ci(x) - \frac{\sin(x)}{x}~~)~~+C}\)

\(\Large\color{blue}{ 2Ci(x)~~- 2(~~Ci(x) - \frac{\sin(x)}{x}~~)~~+C}\)

\(\Large\color{blue}{ 2~(~~~Ci(x)-~~Ci(x) - \frac{\sin(x)}{x}~~)~~+C}\)

\(\Large\color{blue}{ 2~(- \frac{\sin(x)}{x}~~)~~+C}\)

Answer 26

it should be a positive sin(X) .

Answer 27

2sin(X)
------ + C
x

Answer 28

I am not very good at integration... I could have been better if I took calc1, but the highest math course I took is trig -:(

Answer 29

@SolomonZelman you are pretty good at integration by the looks of it :)

Answer 30

could you please teach me the steps on how you did it

Answer 31

yes, myininaya told me I was good at it too, when solved an integral in a couple of steps.

http://openstudy.com/users/solomonzelman#/updates/53dfebc5e4b0ca0415954e5f

Answer 32

Well... I read some staff in a calc 1 book. About it and practiced some probs.

Answer 33

so is this step correct?

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

Answer 34

if its correct what is the next step?

Answer 35

would it be followed by:

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

Answer 36

\[2\int\limits_{}^{}cosx/x-2\int\limits_{}^{}sinx/x^2\]
\[2[\frac{ 1 }{ x }\int\limits_{}^{}cosxdx-\int\limits_{}^{}(d/dx(1/x)*\int\limits_{}^{}cosxdx)-2\int\limits_{}^{}sinx/x^2\]
\[2(1/x)\int\limits_{}^{}cosxdx+2\int\limits_{}^{}(1/x^2)sinxdx-2\int\limits_{}^{}(1/x^2)sinxdx\]
so they are getting calcelled out
\[=\frac{ 2 }{ x }\int\limits_{}^{}cosxdx=(2/x)sinx=\frac{ 2*sinx }{ x }\]
same thing

Answer 37

@SolomonZelman

Answer 38

if u havent learned uv subs then u'll not understand it @Chad123

Answer 39

i have learnt it but in my class the teacher explained very well that we avoid putting x's outside the integral only constants

Answer 40

are u able to grasp it then?

Answer 41

but you havent used any u's in your explanation thats why it is not clear

Answer 42

having things on the screen is different when written on paper

Answer 43

i would like someone to take me through it step by step not write the whole solution

Answer 44

its not just UV
its
LIATE method
L=log
I=inverse
A=algebric
T=trigonometric
E=exponential
i can explain it to u later but i gotta go now sorry

Answer 45

@SolomonZelman are you there?

Answer 46

Yes, I am now here... my friend is moving, so I had to help out.

Answer 47

great im stuck :)

Answer 48

i reposted the question