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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
$\int_{}^{} {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}dx} = $
  • A
    $ - \cot x - 2x + c$
  • B
    $ - 2\cot x - 2x + c$
  • ✓
    $ - 2\cot x - x + c$
  • D
    $ - 2\cot x + x + c$
✓

Answer

Correct option: C.
$ - 2\cot x - x + c$
c
(c) $\int_{}^{} {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}} \,dx = \int_{}^{} {({\rm{cose}}{{\rm{c}}^2}x + {{\cot }^2}x)\,dx} $
$ = \int_{}^{} {(2{\rm{cose}}{{\rm{c}}^2}x - 1)\,dx = - 2\cot x - x + c.} $

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