The integral [_ /6^ /4 dx 2x( ^5x + ^5x ) ] equals :

Question

The integral  $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$  equals :

Accepted Answer

I $=$ $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$

$${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}} $$   Put tan5x $=$ t

$${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}}  = {1 \over {10}}\left( {{\pi  \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$$

Solutions · 1