Find-tan-2tan-2-2-2-tan-2-2-2-n-tan-2-n-

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

Find

\tan θ + 2 \tan 2 θ + 2^{2} \tan 2^{2} θ + \dots + 2^{n} \tan 2^{n} θ

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

f (x) = \tan x + 2 \tan 2 x + \dots + 2^{n} \tan 2^{n} x

\int f (x) d x = - \ln (\cos x \cos 2 x \dots \cos 2^{n} x) + C

\cos x \cos 2 x \dots \cos 2^{n} x = \frac{\sin x \cos x \cos 2 x \dots \cos 2^{n x}}{\sin x}

= \frac{\sin 2^{n + 1} x}{2^{n + 1} \sin x}

\int f (x) d x = \ln 2^{n + 1} + \ln \sin x - \ln \sin 2^{n + 1} x + C

f (x) = \frac{\cos x}{\sin x} - \frac{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