If `U_n=int_0^pi(1-cosnx)/(1-cosx)dx ,` where `n` is positive integer or zero, then show that `U_(n+2)+U_n=2U_(n+1)dot` Hence, deduce that `int_0^(pi/2)(sin^2ntheta)/(sin^2theta)=1/2npidot`

Options

A. `pi//2`
B. `pi`
C. `npi//2`
D. `npi`

Answer

C

Explanation

`U_(n+2)-U_(n+1)=int_(0)^(pi)((1-cos(n+2)x)-(1-cos(n+1)x))/(1-cosx)dx`
`=int_(0)^(pi)(cos(n+1)x-cos(n+2)x)/(1-cosx)`
`=int_(0)^(x)(2sin(n+3/2)x . "sin"x/2)/(2sin^(2)x//2) dx`
`implies U_(n+2)-U_(n+1)=int_(0)^(pi)("sin"(n+3/2)x)/("sin"x/2)dx`.................1
`impliesU_(n+1)-U_(n)=int_(0)^(pi)("sin"(n+1/2)x)/("sin"x/2)dx`.............2
From 1 and 2 we get
`(U_(n+2)-U_(n-1))-(U_(n+1)-U_(n))`
`=int_(0)^(pi)(sin(n+3/2)x-sin(n+1/2)x)/("sin"x/2)dx`
`implies U_(n+2)+U_(n)-2U_(n+1)`
`=int(2cos(n+1)x.sinx//2)/(sinx//2) dx`
`=2int_(0)^(pi)cos(n+1)x dx`
`=2((sin(n+1)x)/(n+1))-(0)^(pi)=0`
`impliesU_(n+2)+U_(n)=2U_(n+1)`
`implies U_(n),U_(n+1),U_(n+2)` are in A.P.
`U_(0)=int_(0)^(pi)(1-1)/(1-cosx)dx=0`
`U_(1)=int_(0)^(pi)(1-cosx)/(1-cosx) dx=pi`
`U_(1)=U_(0)=pi` (common difference)
`:.U_(n)=U_(0)+npi=npi`
Now, `I_(n)=int_(0)^(pi//2) (sin^(2) n theta)/(sin^(2) theta) d theta`
`=int_(0)^(pi//2) (sin^(2) n theta)/(sin^(2)theta) d theta`
`=int_(0)^(pi//2) (1-cos2 n theta)/(1-cos 2theta) d theta=1/2 int_(0)^(pi)(1-cosn x)/(1-cosx) dx`
`impliesI_(n)=1/2npi`