Question

Let $A_n$ be the area bounded by the curve $y=(\tan x{)}^n$ & the lines $x=0,y=0$& $x=\pi /4$. Prove that for $n>2,A_n+A_{n-2}=1/(n-1)$ & deduce that $1/(2n+2)<A_n<1/(2n-1).$

Answer 1

$A_n={\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{2}dx$

$A_{n-2}={\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{n-2}dx$

$A_n+A_{n-2}={\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{n}dx+{\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{n-2}dx$

put $t=\tan x$

$dt={\sec }^{2}xdx$

$\frac{dt}{(1+t^2)}=dx$

$A_2+A_{n-2}={\int }_0^1\frac{t^n}{(1+t^2)}dt+{\int }_0^1\frac{t^{n-2}}{(1+t^2)}dt$

$A_2+A_{n-2}={\int }_0^1[\frac{t^n}{(1+t^2)}+\frac{t^n}{(1+t^2)t^2}]dt$

$A_2+A_{n-2}={\int }_0^1\frac{{(t}^2+1)}{(1+t^2)}\frac{t^n}{t^2}dt$

$={\int }_0^1{t}^{n-2}dt$
$A_2+A_{n-2}={\left[\frac{t^{n-1}}{(n-1)}\right]}_0^1=\frac{1}{(n-1)}$

$u=tanx$ $0<u<1$

$u^{n-2}>u^n$ $u^n>u^{n+2}$

$A_{n-2}>A_n$ $A_n>A_{n-2}$

Adding $A_n$ on both sides Adding $A_n$ on both sides

$A_n+A_{n+2}>2A_n$ $2A_n>A_n+A_{n+2}$

$\frac{1}{(n-1)}>2A_n$ $2A_n>\frac{1}{n+1}$

$A_n<\frac{1}{2n+2}$ $A_n>\frac{1}{2n+2}$

$\frac{1}{2n+2}<A_n<\frac{1}{2n-1}$