InterviewSolution
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InterviewSolution
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If U_(n)=int_(0)^(pi)(1-cosnx)/(1-cosx)dx where n is positive integer of zero, then The value of U_(n) is |
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Answer» `pi//2` `=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` |
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