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QJEEM19Ma8108: JEE Main 2019 Mathematics - Class 12 JEEM19Ma8108 Solution: The value of      where [t] denotes the…

The value of    ∫−π/2π/2dx[x]+[sin⁡x]+4,\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} ,   where [t] denotes the greatest integer less than or equal to t, is

Multiple Choice Options

  • A 112(7π−5){1 \over {12}}\left( {7\pi - 5} \right)
  • B 112(7π+5){1 \over {12}}\left( {7\pi + 5} \right)
  • C 310(4π−3){3 \over {10}}\left( {4\pi - 3} \right)
  • D 320(4π−3){3 \over {20}}\left( {4\pi - 3} \right)

Step-by-Step Verified Solution

I=∫−π2π2dx[x]+[sin⁡x]+4{\rm I} = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}}

=∫−π2−1dx−2−1+4+∫−10dx−1−1+4= \int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over { - 2 - 1 + 4}}} + \int\limits_{ - 1}^0 {{{dx} \over { - 1 - 1 + 4}}}

                 +∫01dx0+0+4+∫1π2dx1+0+4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_0^1 {{{dx} \over {0 + 0 + 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over {1 + 0 + 4}}} }

∫−π2−1dx1+∫−10dx2+∫01dx4+∫1π2dx5\int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over 1} + \int\limits_{ - 1}^0 {{{dx} \over 2}} } + \int\limits_0^1 {{{dx} \over 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over 5}}

(−1+π2)+12(0+1)+14+15(π2−1)\left( { - 1 + {\pi \over 2}} \right) + {1 \over 2}\left( {0 + 1} \right) + {1 \over 4} + {1 \over 5}\left( {{\pi \over 2} - 1} \right)

−1+12+14−15+π2+π10- 1 + {1 \over 2} + {1 \over 4} - {1 \over 5} + {\pi \over 2} + {\pi \over {10}}

−20+10+5−420+6π10{{ - 20 + 10 + 5 - 4} \over {20}} + {{6\pi } \over {10}}

−920+3π5{{ - 9} \over {20}} + {{3\pi } \over 5}

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