Question

evaluate

Answer 1

\[\int\limits_{0}^{1} \frac{ dx }{ \sqrt{1-x^m} }\]

Answer 2

\(m\)? lol.

Answer 3

yes

Answer 4

m is right

Answer 5

Is this a homework thing?

Answer 6

no
i had this thing in question paper

Answer 7

its 1 of the questions i couldn't solve due to lots of absence in the classroom

Answer 8

I'm really not sure... look it up in an integral table...
I could solve it form \(m=1, m=2\)

Answer 9

\(m=0\) results in division by \(0\).

Answer 10

will there be any formula for
\[1^m-x^m\]

Answer 11

i guess not :/

Answer 12

@Hero @Directrix @mathslover

Answer 13

There well it factors it a bit...

Answer 14

@RadEn

Answer 15

can you give a hint

Answer 16

i am clueless about this 1
@electrokid check this when you come

Answer 17

I know that \[
\sum^n x^k = \frac{1-x^n}{1-x}
\]

Answer 18

~_~" cant see the latex

Answer 19

\[\Large
\sum^{n-1}_{k=0} x^k = \frac{1-x^n}{1-x}
\]

Answer 20

:/
some error with server maybe
i wil check this in a while

Answer 21

I don't know how this would help, it's just some random thing.

Answer 22

Why don't you just do the MacLaurin series?

Answer 23

umm
letme check

Answer 24

is that same of tylor series?

Answer 25

It's the Taylor series centered at \(0\)

Answer 26

I see

Answer 27

~_~" clueless again

Answer 28

Don't know how to do the MacLaurin?

Answer 29

how do I aply it here?

Answer 30

\[\Large
f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n
\]
`f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n`

Answer 31

They're really easy to integrate, dude.

Answer 32

arey I know the formula

Answer 33

and I cant see the latex again
wtf
wait I will change the browser

Answer 34

Well, consider :\[\Large
\int \sum^\infty c_nx^ndx = \sum^\infty \frac{c_nx^{n+1}}{n+1}
\]

Answer 35

`\int \sum^\infty c_nx^ndx = \sum^\infty \frac{c_nx^{n+1}}{n+1}`

Answer 36

~_~" is something wrong with server
I am luking at lateX like \int \sum^\infty c_nx^ndx = \sum^\infty \frac{c_nx^{n+1}}{n+1}

Answer 37

I will check this later
sorry

Answer 38

No nothing wrong in ur server @aajugdar :
I am looking the same :
\[\int \sum^\infty c_nx^ndx = \sum^\infty \frac{c_nx^{n+1}}{n+1}\]

Now it is ok actually @wio forgot to put `\[\]` there.

Answer 39

oh okay
I still don't get it anyway @wio

Answer 40

put u= 1-x^m

Answer 41

x= (1-u)^(1/m)
dx = -1/m *(1-u)^(1/m-1)du

Answer 42

just trying....

Answer 43

it came in the form of beta -gamma functions :P

Answer 44

\(
{\beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt
\)

Answer 45

\[
let x^m=u\implies mx^{m-1}dx=du\implies dx=\frac{1}{mu^{1-m\over m}}du\\
dx=\frac{u^{m-1\over m}}{m}du
\]
limits do not change
\[
I=\int_0^1(1-u)^{-1\over2}\frac{u^{m-1\over m}}{m}du\\
I={1\over m}\int_0^1(1-u)^{1\over2}u^{m-1\over m}du\\
\]
the integral is a Beta function http://en.wikipedia.org/wiki/Beta_function
\[
I=\frac{B(p,q)}{m}=\frac{B({m-1\over m},-{1\over2})}{m}\\
I=\frac{\Gamma({m-1\over m})\Gamma(-{1\over2})}{m\Gamma({m-1\over m}-{1\over2}+1)}
\]

Answer 46

correction...
\[
x=u^{1\over m}\implies dx={1\over m}u^{1-m\over m}du
\]

Answer 47

@electrokid why limits do not change???

Answer 48

okk okk got it sorry..:)

Answer 49

yay
thanks ^^ @electrokid

Answer 50

what if
\[
x^m=\sin^2\theta\implies x=\sin^{2\over m}\theta\implies dx={2\over m}\sin^{2-m\over m}\theta\cos\theta d\theta\\
\]
when \(x=0\), \(\theta=0\) and when \(x=1\), \(\theta={\pi\over2}\)
\[
I={2\over m}\int_0^{\pi\over2}\frac{1}{\cos\theta}\sin^{2-m\over m}\theta d\theta\\
I={2\over m}\int_0^{\pi\over2}[\cos\theta]^{-1}\sin^{2-m\over m}\theta d\theta\\
\]

Answer 51

@aajugdar I cant go from there. :(

Answer 52

isn't that another form of beta function.....

Answer 53

yes.. @hartnn look under properties on wiki page

Answer 54

i know it is....i was asking you because you couldn't go ahead...?

Answer 55

so how does it go from there?

Answer 56

help me goku ~_~

Answer 57

wait..i am checking his work...

Answer 58

okay

Answer 59

i forgot to take bath o_o brb

Answer 60

doesn't seem correct :P
how did the exponent became 1/2 from -1/2 ?

Answer 61

\(I=\int_0^1(1-u)^{-1\over2}\frac{u^{m-1\over m}}{m}du\\
I={1\over m}\int_0^1(1-u)^{\color{red}{\huge "-"}1\over2}u^{m-1\over m}du\\\)

Answer 62

also, the exponents are -1/2 and (2-m)/m
so, 1/2-1 and 2/m-1
so, the parameter in beta function should be
B(1/2,2/m)
right ? @electrokid ?

Answer 63

@aajugdar nothing like a good math problem to break the day. thanks. :)

Answer 64

@hartnn yes. looks like it

Answer 65

but, assuming m > 2, the second term becomes less than "1"

Answer 66

do we know the domain of "m"?

Answer 67

hmm
you are welcome
but i got doomed in exam :( :D

Answer 68

no
domain of m is not given
but for 0 its not defined so m>0>m

Answer 69

so,
\[
I={1\over m}B\left({1\over2},{2\over m}\right)={1\over m}\frac{\Gamma({1\over2})\Gamma({2\over m})}{\Gamma({1\over2}+{2\over m})}
\]
an interesting approach:
Express the integral as Reimann's sums?
\[
I=\lim_{n\to\infty}\sum_{k=1}^n\frac{x_k-x_{k-1}}{\sqrt{1-x_k^m}}\qquad x_0=0,x_n=1
\]

Answer 70

~_~ cant it be solvd without involving series

Answer 71

Anyone for the simpsonn's rule ?