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$.
(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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