JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 5)
The value of the definite integral $$\int\limits_{\pi /24}^{5\pi /24} {{{dx} \over {1 + \root 3 \of {\tan 2x} }}} $$ is :
$${\pi \over 3}$$
$${\pi \over 6}$$
$${\pi \over {12}}$$
$${\pi \over {18}}$$
Explanation
Let $$I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\cos 2x)}^{1/3}}} \over {{{(\cos 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx} $$ ...... (i)
$$ \Rightarrow I = \int\limits_{\pi /2}^{5\pi /24} {{{{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}} \over {{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}} + {{\left( {\sin \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}}}} dx\left\{ {\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } } \right\}$$
So, $$I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\sin 2x)}^{1/3}}} \over {{{(\sin 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx} $$ ..... (ii)
Hence, $$2I = \int\limits_{\pi /24}^{5\pi /24} {dx} $$ [(i) + (ii)]
$$ \Rightarrow 2I = {{4\pi } \over {24}} \Rightarrow I = {\pi \over {12}}$$
$$ \Rightarrow I = \int\limits_{\pi /2}^{5\pi /24} {{{{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}} \over {{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}} + {{\left( {\sin \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}}}} dx\left\{ {\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } } \right\}$$
So, $$I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\sin 2x)}^{1/3}}} \over {{{(\sin 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx} $$ ..... (ii)
Hence, $$2I = \int\limits_{\pi /24}^{5\pi /24} {dx} $$ [(i) + (ii)]
$$ \Rightarrow 2I = {{4\pi } \over {24}} \Rightarrow I = {\pi \over {12}}$$
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