A) \[\frac{1}{\log a}{{\sin }^{-1}}{{a}^{x}}+c\]
B) \[{{\sin }^{-1}}{{a}^{x}}+c\]
C) \[\frac{1}{\log a}{{\cos }^{-1}}{{a}^{x}}+c\]
D) \[{{\cos }^{-1}}{{a}^{x}}+c\]
Correct Answer: A
Solution :
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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