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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{{\rm{cose}}{{\rm{c}}^2}x}}{{1 + \cot x}}dx = } $
  • A
    $\log (1 + \cot x) + c$
  • ✓
    $ - \log (1 + \cot x) + c$
  • C
    $\frac{1}{{2{{(1 + \cot x)}^2}}} + c$
  • D
    None of these
✓

Answer

Correct option: B.
$ - \log (1 + \cot x) + c$
b
(b) Put $1 + \cot x = t \Rightarrow {\rm{cose}}{{\rm{c}}^2}x\,dx = - dt,$ then
$\int_{}^{} {\frac{{{\rm{cose}}{{\rm{c}}^2}x}}{{1 + \cot x}}\,dx} = - \int_{}^{} {\frac{1}{t}\,dt = - \log t + c} = - \log (1 + \cot x) + c$.

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