# Thread: integral

1. ## integral

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

2. 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.
$\displaystyle \int_0^2 \frac{x^5}{(x^2+1)^4} dx = \int_0^2 \frac{x^5}{(x^2+1)^3} \cdot \frac{1}{x^2+1} dx$

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

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

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

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