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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{{dx}}{{x\sqrt {1 - {{(\log x)}^2}} }} = } $
  • A
    ${\cos ^{ - 1}}(\log x) + c$
  • B
    $x\log (1 - {x^2}) + c$
  • ✓
    ${\sin ^{ - 1}}(\log x) + c$
  • D
    $\frac{1}{2}{\cos ^{ - 1}}(\log x) + c$
✓

Answer

Correct option: C.
${\sin ^{ - 1}}(\log x) + c$
c
(c) Put $\log x = t \Rightarrow \frac{1}{x}\,dx = dt,$ then it reduces to
$\int_{}^{} {\frac{{dt}}{{\sqrt {1 - {t^2}} }} = {{\sin }^{ - 1}}t = {{\sin }^{ - 1}}(\log x) + c.} $

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