QJEEM21Ma1270: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1270 Solution: The value of the definite integral ∫ π / 24 5 π / 24 d x 1 +…

Question

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The value of the definite integral

\int_{π / 24}^{5 π / 24} \frac{d x}{1 + \sqrt[3]{\tan 2 x}}

is :

Accepted Answer

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}}

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