To solve for the constant [tex]\( A \)[/tex] in the integral equation [tex]\( A \int \frac{d x}{\sqrt{a^2-x^2}}=v^2 \)[/tex], let's break down the process step-by-step.
1. Recognize the Standard Integral:
Consider the integral [tex]\(\int \frac{dx}{\sqrt{a^2 - x^2}}\)[/tex]. This is a standard integral that can be found in integral tables or textbooks on calculus. The result of this integral is:
[tex]\[
\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
2. Set Up the Given Equation:
According to the problem, we have:
[tex]\[
A \int \frac{dx}{\sqrt{a^2 - x^2}} = v^2
\][/tex]
Substitute the result of the integral into this equation:
[tex]\[
A \left(\arcsin\left(\frac{x}{a}\right) + C \right) = v^2
\][/tex]
3. Determine the Form of the Solution:
Since [tex]\( A \)[/tex] is a constant that scales the integral result, we need to normalize the integral’s standard result to equal [tex]\( v^2 \)[/tex]. To do this directly and confirm the value of [tex]\( A \)[/tex]:
4. Extract [tex]\( A \)[/tex]:
From the standard integral form, we have:
[tex]\[
A \arcsin\left(\frac{x}{a}\right) = v^2
\][/tex]
Since [tex]\( v^2 \)[/tex] is given as the result, if we let the standard integral value equal [tex]\( v^2 \)[/tex], it implies that [tex]\( A \)[/tex] must be the necessary scaling factor that makes the result of the standard integral exactly equal [tex]\( v^2 \)[/tex].
So, for the integral [tex]\(\arcsin\left(\frac{x}{a}\right) = v^2\)[/tex] to hold true as an intact relation,
[tex]\[
A \arcsin\left(\frac{x}{a}\right) = \arcsin\left(\frac{x}{a}\right)
\][/tex]
confirms that we should have [tex]\( A = 1 \)[/tex].
Therefore, by standardization and comparison, the value of [tex]\( A \)[/tex] is:
[tex]\[
\boxed{1}
\][/tex]
