Question

${\int }_0^{\infty }\frac{dx}{{(x+\sqrt{1+x^2})}^n}=f\left(n\right)$
Find the value if $n=6$, can be expressed as $a/b$ in simplest form, then $b-a=?$

Answer 1

Let $I={\int }_0^{\infty }\frac{dx}{{(x+\sqrt{1+x^2})}^n}$
Substitute $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$
$I={\int }_0^{\pi /2}\frac{{\sec }^2\theta d\theta }{{(\tan \theta +\sec \theta )}^n}={\int }_0^{\pi /2}\frac{{\cos }^{n-2}\theta d\theta }{{(1+\sin \theta )}^n}$
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_0^{\pi /2}\frac{{\sin }^{n-2}\theta d\theta }{{(1+\cos \theta )}^n}={\int }_0^{\pi /2}\frac{{\left(2\sin \theta /2\cos \theta /2\right)}^{n-2}}{{\left(2{\cos }^2\theta /2\right)}^n}d\theta$
$=\frac{1}{4}{\int }_0^{\pi /2}\frac{{\sin }^{n-2}\theta /2}{{\cos }^{n+2}\theta /2}d\theta =\frac{1}{4}{\int }_0^{\pi /2}\frac{{\sin }^{n-2}\theta /2}{{\cos }^{n-2}\theta /2}d\theta .\frac{1}{{\cos }^4\theta /2}d\theta$
$I=\frac{1}{4}{\int }_0^{\pi /2}{\tan }^{n-2}\theta (1+{\tan }^2\frac{\theta }{2}).{\sec }^2\frac{\theta }{2}.{\sec }^2\frac{\theta }{2}d\theta$
Substitute $\tan \frac{\theta }{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{\theta }{2}d\theta =dt$
$I=\frac{1}{4}{\int }_0^1{t}^{n-2}(1+t^2).2dt=\frac{1}{2}{\int }_0^1(t^{n-2}+t^n)dt$
$=\frac{1}{2}{[\frac{t^{n-1}}{n-1}+\frac{t^{n+1}}{n+1}]}_0^1=\frac{1}{2}[\frac{1}{n-1}+\frac{1}{n+1}]=\frac{n}{n^2-1}$
Hence $b=35,a=6$
Therefore $b-a=29$