Question

$\int \frac{a^{x/2}}{\sqrt{a^{-x}-a^x}}\mathrm{dx}$ is equal to

• (A) $\frac{1}{\log a}{\sin }^{-1}\left(a^x\right)+c$
• (B) $\frac{1}{\log \mathrm{a}}{\tan }^{-1}\left({\mathrm{a}}^x\right)+\mathrm{c}$
• (C) $2\sqrt{a^{-x}-a^x}+c$
• (D) $\log (a^x-1)+c$

Answer 1

The Correct Option is A

Explanation:
Let $I=\int \frac{d^{x/2}}{\sqrt{a^{-x}-a^x}}dx$$=\int \frac{a^x}{\sqrt{1-a^{2x}}}dx$Let $a^x=t\Rightarrow d^x\log adx=dt$$\begin{array}{rl}\therefore & =\int \frac{dt}{\sqrt{1-t^2}}\cdot \frac{1}{\log a}\\ & =\frac{1}{\log a}{\sin }^{-1}(t)+c\\ & =\frac{1}{\log a}{\sin }^{-1}\left(a^x\right)+c\end{array}$