intlimits_{frac{1}{2}}^2 {frac{1}{x}{{tan }^{ | vedclass

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(C) Let $I = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}{{	an }^{2015}}\left( {x - \frac{1}{x}} ight)dx} $. Substitute $x = \frac{1}{t}$,then $dx = -\

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(C) Let $I = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}{{	an }^{2015}}\left( {x - \frac{1}{x}} ight)dx} $. Substitute $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$. When $x = \frac{1}{2}$,$t = 2$. When $x = 2$,$t = \frac{1}{2}$. So,$I = \int\limits_2^{\frac{1}{2}} {t \cdot {{	an }^{2015}}\left( {\frac{1}{t} - t} ight) \cdot \left( -\frac{1}{t^2} ight) dt} $. $I = \int\limits_{\frac{1}{2}}^2 {\frac{1}{t} \cdot {{	an }^{2015}}\left( {-(t - \frac{1}{t})} ight) dt} $. Since $	an(- 	heta) = -	an(	heta)$,we have $	an^{2015}(-(t - \frac{1}{t})) = -	an^{2015}(t - \frac{1}{t})$. Thus,$I = - \int\limits_{\frac{1}{2}}^2 {\frac{1}{t} {{	an }^{2015}}\left( {t - \frac{1}{t}} ight) dt} = -I$. $2I = 0 \Rightarrow I = 0$.

$\int\limits_{\frac{1}{2}}^2 {\frac{1}{x}{{\tan }^{2015}}\left( {x - \frac{1}{x}} \right)dx} $ is equal to

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