what is integral of (cos x) / x ?

- anonymous

what is integral of (cos x) / x ?

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- schrodinger

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- anonymous

What integration techniques do you have right now?

- anonymous

techniques? like?

- anonymous

u-substitution
parts
trig substitution
etc.

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## More answers

- anonymous

i just wanna find out derivative of cosx/x between 3 pi/2 and pi/2

- anonymous

and then the integral of cos x/x

- anonymous

The derivative would be the slope of the tangent line, the integral is the area of the region. Two different (yet related) things.

- anonymous

i have tried finding out integral of cos x/x but it was an infinite loop!!! need an answer for this plz

- anonymous

- anonymous

ntegration by parts is a heuristic rather than a purely mechanical process for solving integrals; given a single function to integrate, the typical strategy is to carefully separate it into a product of two functions ƒ(x)g(x) such that the integral produced by the integration by parts formula is easier to evaluate than the original one.

- anonymous

So it sounds like you tried parts. Which is what I would do. The integral ends up being: \[[\ln(x)\cos(x)+\sin(x)/x]/2 + c\]

- anonymous

uh huh.

- anonymous

how did u get that answer? can u plz explain the steps..

- anonymous

I set u = cos(x) and dv = dx/x and solved for du and v.
Then I put in uv - ∫v du and had to use parts again using u = sin(x) and dv = ln(x)dx and solved for du and v again.
This time I ended up with ln(x)cos(x)+sin(x)/x - ∫cos(x)/x dx = ∫cos(x)/x dx (the original problem) then I added ∫cos(x)/x dx to both sides and divide by two.

- anonymous

hmm..so derv of cos x =-sinx and intg(dv) =1/x dx right?

- anonymous

du = -sin(x)dx and v = ln(x) for the first parts.

- anonymous

the formula is ,
u int (v)- int{d/dx(u) . int(v)}dx right?

- anonymous

∫u dv = u*v - ∫v du is what integration by parts states.

- anonymous

intr(u) dv or integral (uv)?

- anonymous

dont we ve any formula as "something / somethig" in intergral as we ve for derivatives?

- anonymous

This integral cannot be represented in terms of transcendental and algebraic functions. You can only represent it with series.

- anonymous

if we have lower limit as pi/2 and upper one as 3 pi/2 then?

- anonymous

as the sin terms become all zeroes with pi/2 's

- anonymous

It can't be done with any method other than series, even if you constrained the limits.

- anonymous

im sry the cos terms

- anonymous

ohh..isnt there any other way out?

- anonymous

Not from what I researched online. You may be able to use power series. However, u-substitution, backwards u-substitution, integration by parts, partial fractions, etc. will not solve this integral.
If I could remember power series from three years ago I could help, but I don't remember :(

- anonymous

ohh :(...anyways thanks

- anonymous

gotta findout frm somewhere else then

- anonymous

thanks Tbates and jshowa

- anonymous

hey can u just temme how do i join the group chat here??

- anonymous

i have got lot many questions to ask....!!!

- anonymous

anybody there?????

- anonymous

I'm not sure what you do, I tried typing something and it didn't work for me. You can just ask another question.

- anonymous

hmm...

- anonymous

so in such a case we get loops isnt it?

- anonymous

even if got the upper n lower limits for that

- anonymous

can anyone answer this?

- anonymous

I'm not sure, however, I intend on answering this question once I refresh myself with series. So, check back in a day or so.
The limits really don't matter because you still have to evaluate the integral before you use them. And with integration by parts, you will continually get integrals that must be done with parts and it will never end.
You must either construct a power series or Taylor series of cosx/x and estimate the integral.

- anonymous

yeah..that actually does go on n on..

- anonymous

n theres one more thing

- anonymous

n theres one more thing

- anonymous

tangent is a plane and normal is a line on that plane..is this staement correct?

- anonymous

anyone there?

- anonymous

integration by parts would do it

- anonymous

make 1/x your dv and and your u cos(x)

- anonymous

hmm...and how wud the final answer look like?

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