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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int_{1}^{6\pi}([sec^{-1}]+[cot^{-1}])dx$ is equal to       (where $[.]$ denotes greatest integer function)
  • A
    $12\pi-sec1$
  • B
    $6\pi-cot1$
  • C
    $6\pi-cot1-sec1$
  • ✓
    $6\pi-sec1$
✓

Answer

Correct option: D.
$6\pi-sec1$
d
$\int\limits_1^{6\pi } {\left( {\left[ {{{\sec }^{ - 1}}x} \right] + \left[ {{{\cot }^{ - 1}}x} \right]} \right)} dx$

$ = \int\limits_1^{6\pi } {\left[ {{{\sec }^{ - 1}}x} \right]}  + \int\limits_1^{6\pi } {\left[ {{{\cot }^{ - 1}}x} \right]dx} $

$ = \int\limits_1^{\sec 1} {0.dx}  + \int\limits_{\sec 1}^{6\pi } {1.dx + 0 = } 6\pi  - \sec 1$

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