Question Number 1018 by prakash jain last updated on 15/May/15

Find  tan θ+2tan 2θ+2^2 tan 2^2 θ+...+2^n tan 2^n θ

Answered by prakash jain last updated on 16/May/15

f(x)=tan x+2tan 2x+...+2^n tan 2^n x  ∫f(x)dx=−ln (cos x cos 2x ...cos 2^n x)+C  cos x cos 2x ...cos 2^n x=((sin xcos xcos 2x...cos 2^(nx) )/(sin x))  =((sin 2^(n+1) x)/(2^(n+1) sin x))  ∫f(x)dx=ln 2^(n+1) +ln sin x−ln sin 2^(n+1) x+C  f(x)=((cos x)/(sin x))−((2^(n+1) cos 2^(n+1) x)/(sin 2^(n+1) x))  f(x)=cot x−2^(n+1) cot 2^(n+1) x