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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{{{e^x}\;dx}}{{\sqrt {1 - {e^{2x}}} }} = } $
  • A
    ${\cos ^{ - 1}}({e^x}) + c$
  • ✓
    $ - {\cos ^{ - 1}}({e^x}) + c$
  • C
    ${\cos ^{ - 1}}({e^{2x}}) + c$
  • D
    $\sqrt {1 - {e^{2x}}} + c$
✓

Answer

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

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