# problem of Integral proof

• Nov 5th 2009, 12:30 AM
kin
problem of Integral proof
• Nov 5th 2009, 12:39 AM
chisigma
You have to take into account that for $|x|<1$ is...

$\frac{x^{4}\cdot (1-x)^{4}}{1-x^{2}}= (1-x)^{4}\cdot (x^{4} + x^{6} + x^{8} + \dots)$

Kind regards

$\chi$ $\sigma$
• Nov 5th 2009, 01:02 AM
kin
Quote:

Originally Posted by chisigma
You have to take into account that for $|x|<1$ is...

$\frac{x^{4}\cdot (1-x)^{4}}{1-x^{2}}= (1-x)^{4}\cdot (x^{4} + x^{6} + x^{8} + \dots)$

Kind regards

$\chi$ $\sigma$

it's $\frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}$ not (1-x),,,,,,,,,
• Nov 5th 2009, 01:32 AM
chisigma
Quote:

Originally Posted by kin
it's $\frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}$ not (1-x)...

Very sorry!(Crying)...

... in such a case is...

$\frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}= (1-x)^{4}\cdot (x^{4} - x^{6} + x^{8} -...)$, $|x|<1$

Kind regards

$\chi$ $\sigma$
• Nov 5th 2009, 02:40 AM
Laurent
Quote:

Originally Posted by kin

how to solve (b) ......?

It seems that you're not expected to use a) to evaluate the integral. The simplest way anyway is probably to expand the numerator, and perform the Euclidian division of $X^4(1-X)^4$ by $1+X^2$ (like with integers, you know?), and get something like $X^4(1-X)^4=Q(X)(1+X^2)+aX+b$ hence your integral equals $\int_0^1 Q(x)dx + a\int_0^1 \frac{x}{1+x^2}dx + b\int_0^1\frac{dx}{1+x^2}$. The first term is easy, the second term has antiderivative $\frac{1}{2}\log(1+x^2)$, and the third one relates to arctan.

For the second part, you probably have to write $1\leq 1+x^2\leq 2$ hence $\frac{1}{2}x^4(1-x)^4\leq \frac{x^4(1-x)^4}{1+x^2}\leq x^4(1-x)^4$ and integrate.
• Nov 5th 2009, 04:45 AM
kin
Quote:

Originally Posted by chisigma
Very sorry!(Crying)...

... in such a case is...

$\frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}= (1-x)^{4}\cdot (x^{4} - x^{6} + x^{8} -...)$, $|x|<1$

Kind regards

$\chi$ $\sigma$

but i still dont understand your point...
how does it help to solve the integral..?
• Nov 5th 2009, 04:48 AM
kin
Quote:

Originally Posted by Laurent
It seems that you're not expected to use a) to evaluate the integral. The simplest way anyway is probably to expand the numerator, and perform the Euclidian division of $X^4(1-X)^4$ by $1+X^2$ (like with integers, you know?), and get something like $X^4(1-X)^4=Q(X)(1+X^2)+aX+b$ hence your integral equals $\int_0^1 Q(x)dx + a\int_0^1 \frac{x}{1+x^2}dx + b\int_0^1\frac{dx}{1+x^2}$. The first term is easy, the second term has antiderivative $\frac{1}{2}\log(1+x^2)$, and the third one relates to arctan.

For the second part, you probably have to write $1\leq 1+x^2\leq 2$ hence $\frac{1}{2}x^4(1-x)^4\leq \frac{x^4(1-x)^4}{1+x^2}\leq x^4(1-x)^4$ and integrate.

thank u very much ,
this is a straightforward method....
are there any ways of combining result of (a) to solve it quickly...?
• Nov 5th 2009, 06:21 AM
debanik1
which book are these problems from?
• Nov 5th 2009, 06:37 AM
kin
Quote:

Originally Posted by debanik1
which book are these problems from?

they are not from books....just from my tutorial notes...

anyway, i have solved it.

thanks for help , guys...(Hi)