The value of the definite integral $\int\limits_{\pi /24}^{5\pi /24} {{{dx} \over {1 + \root 3 \of {\tan 2x} }}}$ is :

Let $$I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\cos 2x)}^{1/3}}} \over {{{(\cos 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx} $$ ...... (i)<br><br>$$ \Rightarrow I = \int\limits_{\pi /2}^{5\pi /24} {{{{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}} \over {{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}} + {{\left( {\sin \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}}}} dx\left\{ {\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } } \right\}$$<br><br>So, $$I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\sin 2x)}^{1/3}}} \over {{{(\sin 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx} $$ ..... (ii)<br><br>Hence, $2I = \int\limits_{\pi /24}^{5\pi /24} {dx}$ [(i) + (ii)]<br><br>$\Rightarrow 2I = {{4\pi } \over {24}} \Rightarrow I = {\pi \over {12}}$