Verified Solution • 2025–26 Session moderate

QJEEM19Ma4ae8: JEE Main 2019 Mathematics - Class 12 JEEM19Ma4ae8 Solution: The integral is equal to :

The integral ∫π/6π/3sec⁡2/3xcos⁡ec4/3xdx\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}} x\cos e{c^{4/3}}xdx is equal to :

Multiple Choice Options

  • A 353−313{3^{{5 \over 3}}} - {3^{{1 \over 3}}}
  • B 356−323{3^{{5 \over 6}}} - {3^{{2 \over 3}}}
  • C 343−313{3^{{4 \over 3}}} - {3^{{1 \over 3}}}
  • D 376−356{3^{{7 \over 6}}} - {3^{{5 \over 6}}}

Step-by-Step Verified Solution

∫π/6π/3sec⁡2/3x cosec4/3x dx\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}x\,{{{\mathop{\rm cosec}\nolimits} }^{4/3}}x\,dx}

=∫sec⁡2xtan⁡4/3xdx= \int {{{{{\sec }^2}x} \over {{{\tan }^{4/3}}x}}dx}

Let tan x = t, sec2 x dx = dt

=∫dtt4/3= \int {{{dt} \over {{t^{4/3}}}}}

I=−3(t−1/3)I = - 3\left( {{t^{ - 1/3}}} \right)

⇒(−3(tan⁡x)−13)π/6π/3\Rightarrow \left( { - 3{{\left( {\tan x} \right)}^{{{ - 1} \over 3}}}} \right)_{\pi /6}^{\pi /3}

⇒3(313−1316)\Rightarrow 3\left( {{3^{{1 \over 3}}} - {1 \over {{3^{{1 \over 6}}}}}} \right) = 37/6 - 35/6

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