Let `A_(n)` be the area bounded by the curve `y=(tan x)^(n)` and the lines `y=0, and x=(pi)/(4)." Prove that for "ngt2, A_(n)+A_(n-2)=(1)/(n-1)" and deduce that "(1)/(2n+2)ltA_(n)lt(1)/(2n-2).`

Explanation

We have, ` A_(n)= int_(0)^(pi//4)(tanx)^(n)dx `
Since, `0 lt tanx lt 1, " when " 0 lt x lt pi//4 `
We have, `0 lt (tanx)^(n+1) lt (tanx)^(n) " for each " n in N `
` rArr int_(0)^(pi//4)(tanx)^(n+1)dx lt int_(0)^(pi//4)(tanx)^(n)dx `
` rArr A_(n+1) lt A_(n) `
Now,for `n gt 2`
` A_(n)+A_(n+2)=int_(0)^(pi//4)[(tanx)^(n)+(tanx)^(n+2)]dx `
` =int_(0)^(pi//4)(tanx)^(n)(1+tan^(2)x)dx `
` (##41Y_SP_MATH_C13_E02_041_S01.png" width="80%">
`=int_(0)^(pi//4)(tanx)^(n)sec^(2)xdx `
`=[(1)/((n+1))(tanx)^(n+1)]_(0)^(pi//4)`
`=(1)/((n+1))(1-0)=(1)/(n+1) `
Since, ` A_(n+2) lt A_(n+1) lt A_(n), `
then `A_(n)+A_(n+2) lt 2 A_(n)`
` rArr (1)/(n+1) lt 2 A_(n) `
`rArr (1)/(2n+2) lt A_(n) " "...(i)`
Also, for ` n gt 2A_(n)+A_(n) lt A_(n)+A_(n-2)=(1)/(n-2)`
` rArr 2A_(n) lt (1)/(n-1) `
`rArr A_(n) lt (1)/(2n-2) " "...(ii)`
From Eqs. (i) and (ii),` (1)/(2n+2) lt A_(n) lt (1)/(2n-2) `