# Thread: problem of Integral proof

1. ## problem of Integral proof

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

2. 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

$\chi$ $\sigma$

3. 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),,,,,,,,,

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

... 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

$\chi$ $\sigma$

5. 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.

6. Originally Posted by chisigma
Very sorry!...

... 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..?

7. 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...?

8. which book are these problems from?

9. 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...