If `I_(n)= int_(0)^(4) x dx` then what is `I_(n) + I_(n-2)` equal to ?

Options

A. `1/n`
B. `(1)/((n-1))`
C. `n/((n-1))`
D. `(1)/((n-2))`

Answer

B

Explanation

`I = int_(0)^(pi) (dx)/(1+2sin^(2)x)= 2int_(0)^(pi//2) (dx)/(1+2sin^(2)x)`
` = 2int_(0)^(pi//2) (sec^(2)xdx)/(sec^(2)x+2tan^(2)x) = 2int_(0)^(pi//2) (sec^(2) xdx)/(1+3tan^(2)x)`
Put `tan x = trArr sec^(2) xdx = dt`
When `x = 0, t = 0` and when `x=pi/4= t = 1`
`I_(n) + I_(n-2) = int_(0)^(1) t^(n-2) dt`
`= (t^(n-2 + 1))/(n-2 + 1)"|"_(0)^(1) = (t^(n-1))/(n-1)|_(0)^(1) = (1)/(n-1)[1-0] = (1)/(n-1)`