A) \[\frac{1}{\log a}{{\sin }^{-1}}({{a}^{x}})+c\]
B) \[\frac{1}{\log a}{{\tan }^{-1}}({{a}^{x}})+c\]
C) \[2\sqrt{{{a}^{-x}}-{{a}^{x}}}+c\]
D) \[\log ({{a}^{x}}-1)+c\]
E) \[{{\sin }^{-1}}({{a}^{x}})+c\]
Correct Answer: A
Solution :
Let \[I=\int{\frac{{{a}^{x/2}}}{\sqrt{{{a}^{-x}}-{{a}^{x}}}}}dx\] \[=\int{\frac{{{a}^{x}}}{\sqrt{1-{{a}^{2x}}}}}dx\] Let \[{{a}^{x}}=t\Rightarrow {{a}^{x}}\log a\,dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{\sqrt{1-{{t}^{2}}}}.\frac{1}{\log a}}\] \[=\frac{1}{\log a}{{\sin }^{-1}}(t)+c\] \[=\frac{1}{\log a}{{\sin }^{-1}}({{a}^{x}})+c\]
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