Sure, let's evaluate the integral [tex]\( \int \frac{dx}{\sqrt{2x - x^2}} \)[/tex].
1. Rewriting the integrand:
To simplify the expression inside the square root, we first complete the square for the quadratic [tex]\(2x - x^2\)[/tex].
[tex]\[
2x - x^2 = - (x^2 - 2x)
\][/tex]
To complete the square, we rewrite it as:
[tex]\[
x^2 - 2x = (x - 1)^2 - 1
\][/tex]
Substituting back, we get:
[tex]\[
2x - x^2 = -[(x - 1)^2 - 1] = 1 - (x - 1)^2
\][/tex]
2. Substituting back into the integral:
With the completed square form, the integral becomes:
[tex]\[
\int \frac{dx}{\sqrt{1 - (x - 1)^2}}
\][/tex]
3. Recognizing the standard form:
The integrand is now in a standard form that matches the arcsine function. Recall the integral:
[tex]\[
\int \frac{du}{\sqrt{1 - u^2}} = \arcsin(u) + C
\][/tex]
In our case, [tex]\(u = x - 1\)[/tex].
4. Applying the standard formula:
Using [tex]\( u = x - 1 \)[/tex], the integral becomes:
[tex]\[
\int \frac{dx}{\sqrt{1 - (x - 1)^2}} = \arcsin(x - 1) + C
\][/tex]
5. Final result:
Therefore, the evaluated integral is:
[tex]\[
\int \frac{dx}{\sqrt{2x - x^2}} = \arcsin(x - 1) + C
\][/tex]
Here, [tex]\(C\)[/tex] is the constant of integration.
