integral

**storchfire1X** · May 21st 2007, 01:53 PM

Hi all

The problem is: Show that the value of the integral

int_{0,2} (375x^5)(x^2 +1)^(-4) dx is 2^n for some integer n.

Thank you

**ThePerfectHacker** · May 21st 2007, 07:58 PM

Originally Posted by storchfire1X

Hi all

The problem is: Show that the value of the integral

int_{0,2} (375x^5)(x^2 +1)^(-4) dx is 2^n for some integer n.

Thank you

Let us ignore the 375 factor for now.
$\int_0^2 \frac{x^5}{(x^2+1)^4} dx = \int_0^2 \frac{x^5}{(x^2+1)^3} \cdot \frac{1}{x^2+1} dx$

Let $t=\tan^{-1} x$ thus, $t'=\frac{1}{x^2+1}$ .
And $\sin t = x$ and $\cos t = \frac{1}{\sqrt{x^2+1}}$ , hence, $x^5 = \sin^5 t \mbox{ and }\frac{1}{(x^2+1)^3} = \cos^6 t$

Substitution theorem yields,
$\int_0^{\tan^{-1} 2} \frac{\sin^5 t}{\cos^6 t} dt = \int_0^{\tan^{-1} 2} \frac{(1-\cos^2 t)^2 \sin t}{\cos^6 t}dt$

Let $s=\cos t$ then $s'=-\sin t$

The substitution rule says that,
$-\int_1^{2/\sqrt{5}} \frac{(1-s^2)^2}{s^6} ds = \int_{2/\sqrt{5}}^1 \frac{1-2s^2+s^4}{s^6}ds$
Now show that multiplication by 375 will make this an exponent of 2. The rest is more computational.

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