We are asked to solve the integral $\int \frac{1}{\sqrt{100-8x^2}} dx$.

1. Problem Description

We are asked to solve the integral $\int \frac{1}{\sqrt{100-8x^2}} dx$.

2. Solution Steps

First, we can factor out 100 from the square root to get:

$\int \frac{1}{\sqrt{100(1-\frac{8x^2}{100})}} dx = \int \frac{1}{10\sqrt{1-\frac{8x^2}{100}}} dx = \frac{1}{10} \int \frac{1}{\sqrt{1-\frac{8x^2}{100}}} dx$

Now let $u = \frac{\sqrt{8}x}{10}$. Then $du = \frac{\sqrt{8}}{10} dx$, so $dx = \frac{10}{\sqrt{8}} du$.

Substitute these into the integral:

$\frac{1}{10} \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{10}{\sqrt{8}} du = \frac{1}{10} \cdot \frac{10}{\sqrt{8}} \int \frac{1}{\sqrt{1-u^2}} du = \frac{1}{\sqrt{8}} \int \frac{1}{\sqrt{1-u^2}} du$

We know that $\int \frac{1}{\sqrt{1-u^2}} du = \arcsin(u) + C = \sin^{-1}(u) + C$.

Therefore,

$\frac{1}{\sqrt{8}} \int \frac{1}{\sqrt{1-u^2}} du = \frac{1}{\sqrt{8}} \sin^{-1}(u) + C = \frac{1}{\sqrt{8}} \sin^{-1}(\frac{\sqrt{8}x}{10}) + C$.

3. Final Answer

The final answer is A. $\frac{1}{\sqrt{8}} \sin^{-1}(\frac{\sqrt{8}x}{10}) + C$.