If `I_(n)=int_(0)^(pi//4) tan^(n)x dx, (ngt1` is an integer ), then (a) `I_(n)+I_(n-2)=1/(n+1)` (b) `I_(n)+I_(n-2)=1/(n-1)` (c) `I_(2)+I_(4),I_(6),…….` are in H.P. (d) `1/(2(n+1))ltI_(n)lt1/(2(n-1))`

Options

A. `I_(n)+I_(n-2)=1/(n+1)`
B. `I_(n)+I_(n-2)=1/(n-1)`
C. `I_(2)+I_(4),I_(6),…….` are in H.P.
D. `1/(2(n+1))ltI_(n)lt1/(2(n-1))`

Answer

B, C, D

Explanation

`I_(n)=int_(0)^(pi//4)tan^(n)xdx`
`=int_(0)^(pi//4)tan^(n-2)xtan^(2)xdx`
`=int_(0)^(pi//4)sec^(2)x tan^(n-2)x dx-int_(0)^(pi//4) tan^(n-2) xdx`
`=int_(0)^(1)t^(n-2)dt=I_(n-2)`, where `t=tanx`
`:.I_(n)+I_(n-2)=((t^(n-1))/(n-1))_(0)^(1)=1/(n-1)`
Thus `I_(2)+I_(4)+I_(4)+I_(6)..............` are in H.P.
For `0lt xlt pi//4` we have `0 lt tan^(n-2)x`
So `0ltI_(n)ltI_(n-2)` or `I_(n)+I_(n+2)lt 2I_(n)ltI_(n)+I_(n-2)`
or `1/(n+1)lt 2I_(n) lt 1/(n-1)` or `1/(2(n+1)) lt I_(n) lt 1/(2(n-1))`