Verified Solution • 2025–26 Session moderate

QJEEM07Maaab5: JEE Main 2007 Mathematics - Class 12 JEEM07Maaab5 Solution: The solution for of the equation is

The solution for xx of the equation ∫2xdttt2−1=π2\int\limits_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }} = {\pi \over 2}} is

Multiple Choice Options

  • A 32{{\sqrt 3 } \over 2}
  • B 222\sqrt 2
  • C 22
  • D None

Step-by-Step Verified Solution

∫2xdttt2−1=π2\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}

∴\therefore [sec⁡−1t]2x=π2\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}

[\left[ {} \right. As ∫dxxx2−1=sec⁡−1x\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x ]\left. {} \right]

⇒sec⁡−1x−sec⁡−12=π2\Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}

⇒sec⁡−1x−π4=π2\Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}

⇒sec⁡−1x=π2+π4\Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}

⇒sec⁡−1x=3π4\Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}

⇒x=sec⁡3π4\Rightarrow x = \sec {{3\pi } \over 4}

⇒x=−2\Rightarrow x = - \sqrt 2

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