Question

The value of $\int \frac{10^{x/2}}{\sqrt{10^{-x} - 10^{x}}}dx$ is

• A. $\frac{1}{\log_{e} 10} \sin^{-1} \left(10^{x} \right)+c $
• B. $2\sqrt{10^{-x} + 10^{x} } +c $
• C. $\frac{1}{\log_{e} 10}\sinh^{-1} \left(10x\right)+c$
• D. $\frac{-1}{\log_{e} 10}\sinh^{-1} \left(10x\right)+c$

Answer 1

Answer: (A)

Let $I =\int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x} -10^{x}} } dx = \int \frac{10^{\frac{x}{2}}}{\sqrt{\frac{1-10^{2x}}{10^{x}}}} dx $
$= \int \frac{10^{\frac{x}{2}} .10^{\frac{x}{2}}}{\sqrt{1-\left(10^{x}\right)^{2}}} dx = \int\frac{10^{x} }{\sqrt{1-\left(10^{x}\right)^{2}}}dx $
Put $10^{x} =t \Rightarrow 10^{x} \log10 dx =dt $
$\therefore \, \, I = \frac{1}{\log 10} \int \frac{dt}{\sqrt{1-t^{2}} } = \frac{1}{\log 10}\left(\sin^{-1} t\right)+c $
$= \frac{1}{\log 10}\sin^{-1} \left(10^{x}\right)+c $