Question

Not Fool's problem of the day:
Prove\[\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}dx=\frac{\pi^2}{4}\]

Answer 1

This has already been solved, so you could probably find the answer in my profile, but I wanna see a different approach.

Answer 2

let's check out your profile then ... LOL

Answer 3

intergration by parts and subs cosx=y ???

Answer 4

no, tha would be cheating :(
use your own method

Answer 5

ehh.. don't worry, I have no idea about integrals. I can't do it. No worries ;)

Answer 6

integration by parts is a good start @experimentX

Answer 7

i don't think i will be posting ... if i'ever post ... i will post one or two steps before solution. I am terrible at latex.

Answer 8

Lol I love the name! :D

Answer 9

@FoolForMath sorry, didn't mean to steal your moment :P
a coincidence I assure you!

Answer 10

Lolz sure mate :)

Answer 11

We have
\[I= \int_{0} ^{\pi} \frac{x \times \sin x}{1+\cos^2 x} dx..................(1)\]
We know differential integral property
\[I= \int_a ^b f(x) dx=\int_a ^b (a+b-x) dx\]
so
\[I= \int_{0} ^{\pi} \frac{(\pi-x) \times \sin(\pi- x)}{1+\cos^2 (\pi-x)} dx\]
we get now
\[I= \int_{0} ^{\pi} \frac{(\pi-x) \times \sin x}{1+\cos^2 x} dx..................(2)\]
Add 1 and 2 , we get
\[2I= \int _{0} ^{\pi} \frac{(\pi) \times \sin x}{1+\cos^2 x} dx\]
now substitute cos x= u and solve !!!

Answer 12

a way lot better method than i would have imagined ... everybody has one specialty. i guess for @ffm it smelling numbers.

Answer 13

Lol, I love numbers!

Answer 14

Someone stop this @Fweaa guy !!

Answer 15

@Fweaa please don't spam , post your question only once and not in someone else's question!!!!

Answer 16

Turing bring in the hammer :P

Answer 17

awesome, quite different than Zarkon's approach, which I will now post
and I'l get this Fw whatyever guy ;)

Answer 18

@ash2326: Amazing! funny thing I understand your approach :P

Answer 19

Thanks @FFM

Answer 20

The silly thing is my friends who are not great in mathematics finds integration so easy but I always find it difficult after high school :(

Answer 21

here is Zarkons solution:
http://openstudy.com/users/turingtest#/updates/4f047463e4b075b56651c384
@ash2326 thanks for another approach :)
I do see similarities though

Answer 22

Turing: What's up with the Fweaa??!!

Answer 23

thats what i would have done.
--> the classic way.

Answer 24

Everybody would had solved this except me :P

Answer 25

@TuringTest
I'll continue the solution
\[2I=\pi \int_0 ^{\pi} \frac{\sin x}{1+\cos ^2 x} dx\]
u= cos x
so
du =-dx sin x
x=0 u=1
x=pi u=-1
\[2I=\pi \int _{1}^{-1} \frac{- du}{1+u^2}\]
so we have now
\[2I=\pi \int _{-1} ^{1} \frac{du}{1+u^2}\]
we know for an even function
\[\int_{-a} ^{a} f(x) dx= 2\int_0 ^{a} f(x) dx\]
so
we get
\[2I=2\pi \int_0 ^{1} \frac{du}{1+u^2} du\]
or
\[I= \pi \times (\tan^{-1} 1)= \frac{\pi^2}{4}\]

Answer 26

@ffm you didn't try at all, you would definitely, but the way ash2326 done is genius ... i'll remember it.

Answer 27

sorry, took Fwee is the first I've suspended, so I had to figure out exactly how it works
@ash2326 awesome! quite different from the other way :)

Answer 28

Lol, Mate I was good at it since high school ;) I lost interest after I met number theory and combinatoric in clg. But may be that's an lame excuse :(

Answer 29

you are in different league. I am still struggling with calculus.

Answer 30

Thanks @TuringTest

Answer 31

I just love integrals. This was a fun one.
PS:
@experimentX I don't think you're tagging FoolForMath by writing "@ffm"

Answer 32

Hey guys! I just opened my book and I saw this problem, exactly same approach as ash!

Answer 33

he's here ... besides @ffm is more convenient than using whole name.

Answer 34

Yeah, this approach is taught everywhere in India:D

Answer 35

Sure @experiment!! anything fool, ffm whatever you want :P

Answer 36

Okay I think I got a good problem for you guys :P I would post it in a new thread ;) Ash you are not supposed to use our indigenous trick :P

Answer 37

Yeah:D:D I shouldn't have:P

Answer 38

Accidentally, I discovered that it was one of my mock exam question.. and my teacher did it like that.