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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{{a^x}}}{{\sqrt {1 - {a^{2x}}} }}dx = } $
  • ✓
    $\frac{1}{{\log a}}{\sin ^{ - 1}}{a^x} + c$
  • B
    ${\sin ^{ - 1}}{a^x} + c$
  • C
    $\frac{1}{{\log a}}{\cos ^{ - 1}}{a^x} + c$
  • D
    ${\cos ^{ - 1}}{a^x} + c$
✓

Answer

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

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