Question

Prove that \[\int_0^\infty \frac{x\sin{x}}{1+\cos^2{x}}dx=\frac{\pi^2}{4}\]

Answer 1

I don't know how to do this one. Do I need to use Riemman sums?

Answer 2

Any idea guys?

Answer 3

is that even true?

Answer 4

Yes it is.

Answer 5

Try it on wolframalpha and check.

Answer 6

it seems that the indefinite integral is not elementary function

Answer 7

even wolfram alpha gave up
http://www.wolframalpha.com/input/?i=Integrate%5B%28x*Sin%5Bx%5D%29%2F%281+%2B+Cos%5Bx%5D%5E2%29%2C+%7Bx%2C+0%2C+Infinity%7D%5D

Answer 8

This is bad =/.

Answer 9

How do you know that @Tyifan12879?

Answer 10

because i didnt work it out...

Answer 11

integral of xf(x)dx with lim 0 to b = integral of b/2 f(x)dx. if the function f(x)=f(b-x)

Answer 12

You mean \[\int_0^b xf(x)dx=\int_0^b\frac{b}{2}f(x)dx\]? @binary3i?

Answer 13

if \[f(x)=f(b-x)\]

Answer 14

yes but here b is \[\infty\]

Answer 15

Yes that is a big problem =/

Answer 16

Thinking...

Answer 17

I give up for today. Thank you all!

Answer 18

http://www.analyzemath.com/calculus/limits/squeezing.html

You might be able to use this, though I'm not too sure

Answer 19

Oh man, you think I can get into this school no-data? I'm already posting some of these questions in meta-math.
Somebody tell me how to do these.

Answer 20

mathmate's cooking up something good I hope...

Answer 21

Let
\[I\ =\ \int\limits\limits xsin(x)dx = \sin(x) - x \cos(x) +C\]
which clearly diverges (and oscillates) as x approaches infinity.
The denominator of the integrand,
\[1\le (1+\cos^2(x)) \le 2\]
makes the required integral behave similarly to I above.
This leads me to believe that the given improper integral does not exist.

Is it possible to check if there is a typo in either the integral or the result?

Answer 22

I highly doubt it, this question is one of a series on an entrance exam to a rather prestigious school in Mexico.
http://assets.openstudy.com/updates/attachments/4f03e1ade4b075b566519b63-no-data-1325655819441-cinvestaventranceexamjul2011.pdf
I find it hard to imagine that there are mistakes like this on the test.

Answer 23

Any ideas Zarkon?

Answer 24

did you look at the limits of the question you posted?

Answer 25

Did you notice that the upper limit is pi, and not infinity?
So this integral at least exists!

Answer 26

mathmate is correct with the problem in this thread

Answer 27

I didn't post it, no-data did. That sure makes a lot more sense, but I still don't know how to go about it.

Answer 28

I was letting you know that the above is not the same as the linked problem you posted

Answer 29

The integral from 0 to pi checks with pi^2/4 using numerical integration.
So back to the drawing boards, guys!

Answer 30

Right, I just got the link myself so I'm not familiar with the problems. Should have checked though. I wonder how much time they give you for this test.

Answer 31

at least now wolfram agrees that it's true.
http://www.wolframalpha.com/input/?i=Integrate%5B%28x*Sin%5Bx%5D%29%2F%281+%2B+Cos%5Bx%5D%5E2%29%2C+%7Bx%2C+0%2C+pi%7D%5D

Answer 32

is can be done using integration by parts

Answer 33

as it is?

Answer 34

yes...there is a little trick when evaluating the integral after parts is done once

Answer 35

no...with pi...not infinity

Answer 36

the integral does not converge when the upper limit is infinity

Answer 37

no, I got that bit, mathmate made a convincing argument.

Answer 38

use u=x and dv =the rest of the integrand

Answer 39

ok thanks, I'll give it a try.

Answer 40

Yeah, I don't know how to integrate\[\int\tan^{-1}(\cos x)dx\]is that even right so far?

Answer 41

I mean I know I didn't include the other terms, but it seems to come to this integral

Answer 42

you need the limits of integration

Answer 43

but how can I use them if I have no idea how to integrate that?

Answer 44

I said it is a trick :)

Answer 45

you need them..they are very important

Answer 46

I'm also stuck at \[\int \frac {x sin{x} dx}{1+cos^{2}{x}}=-x tan^{-1}{cos{x}}+\int tan^{-1}{cos{x}} dx\]

Answer 47

I'm not going to sleep tonight :/

Answer 48

you want a hint?

Answer 49

I think I do

Answer 50

So are you saying that to solve this I need to use the limits?

Answer 51

do you know what this would be ...

let f be an odd function

what is

\[\int\limits_{-a}^{a}f(x)dx\]

Answer 52

0

Answer 53

yes

Answer 54

you need to use that

Answer 55

is arctan odd?

Answer 56

it is

Answer 57

no

Answer 58

really?

Answer 59

yes

Answer 60

yes...arctan(cos(x)) is not...but you can fix that :)

Answer 61

of course...

Answer 62

Ok no more hints. This may not get solved by me tonight though.