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QJEEM19Ma66d3: JEE Main 2019 Mathematics - Class 12 JEEM19Ma66d3 Solution: If = m( + n), then m.n is equal to

If ∫0π2cot⁡xcot⁡x+cos⁡ecxdx\int\limits_0^{{\pi \over 2}} {{{\cot x} \over {\cot x + \cos ecx}}} dx = m( π\pi + n), then m.n is equal to

Multiple Choice Options

  • A -1
  • B 1
  • C −12- {1 \over 2}
  • D 12{1 \over 2}

Step-by-Step Verified Solution

∫0π/2cot⁡xcot⁡x+cosec xdx=∫0π/2cos⁡xcos⁡x+1dx\int\limits_0^{\pi /2} {{{\cot x} \over {\cot x + {\mathop{\rm cosec}\nolimits} \,x}}dx = } \int\limits_0^{\pi /2} {{{\cos x} \over {\cos x + 1}}dx}

⇒∫0π/22cos⁡2x2−12cos⁡2x2dx=∫0π/2(1−12sec⁡2x2)dx\Rightarrow \int\limits_0^{\pi /2} {{{2{{\cos }^2}{x \over 2} - 1} \over {2{{\cos }^2}{x \over 2}}}dx = } \int\limits_0^{\pi /2} {\left( {1 - {1 \over 2}{{\sec }^2}{x \over 2}} \right)dx}

⇒(x−tan⁡x2)0π/2=π2−1=12(π−2)\Rightarrow \left( {x - \tan {x \over 2}} \right)_0^{\pi /2} = {\pi \over 2} - 1 = {1 \over 2}\left( {\pi - 2} \right)

mn = 12x−2=−1{1 \over 2}x - 2 = - 1

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