Integral Calculus Question 391
Question: $ \int_{{}}^{{}}{\frac{a^{x}}{\sqrt{1-a^{2x}}}dx=} $
[MNR 1983, 87]
Options:
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 $
Show Answer
Answer:
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 $ .
BETA
