What is the integral:
1/root(2x-x^2)
Sam, how does reposting the same question over and over somehow change the answer that was already given to you ??
To find the integral of \( \frac{1}{\sqrt{2x-x^2}} \), we can first simplify the expression by completing the square.
Step 1: Rewrite the denominator as a perfect square:
\(2x-x^2 = -(x^2-2x) = -((x-1)^2 - 1)\)
Step 2: Simplify further:
\(2x-x^2 = -((x-1)^2 - 1) = -((x-1)^2 - 1^2) = -(x-1)^2 + 1\)
Step 3: Rewrite the original integral in terms of the simplified expression:
\(\int \frac{1}{\sqrt{2x-x^2}} dx = \int \frac{1}{\sqrt{-(x-1)^2+1}} dx\)
Now, we can proceed with the integration.
Step 4: Use a trigonometric substitution to simplify the integral. Let \( x - 1 = \sin(t) \), then \( dx = \cos(t) dt \).
This substitution helps to eliminate the square root in the denominator. Now we need to find the new limits of integration.
When \( x = 0 \), then \( x - 1 = -1 \) and \( \sin(t) = -1 \), so \( t = -\frac{\pi}{2} \).
When \( x = 2 \), then \( x - 1 = 1 \) and \( \sin(t) = 1 \), so \( t = \frac{\pi}{2} \).
Therefore, the definite integral becomes:
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{-(x-1)^2+1}} \cos(t) dt\)
Step 5: Simplify the integrand:
\(\frac{1}{\sqrt{-(x-1)^2+1}}\) becomes \( \frac{1}{\sqrt{\cos^2(t) + 1}} \), which simplifies to \( \frac{1}{\sqrt{2\cos^2(t)}} \), and further simplifies to \( \frac{1}{\sqrt{2}} \sec(t) \).
Step 6: Evaluate the integral:
The integral of \( \frac{1}{\sqrt{2}} \sec(t) \) with respect to \( t \) is \( \frac{1}{\sqrt{2}} \ln|\sec(t) + \tan(t)| \).
Finally, substitute back for \( t \) using the original substitution of \( x-1 = \sin(t) \):
\( \frac{1}{\sqrt{2}} \ln|\sec(t) + \tan(t)| \Bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \)
Evaluate the limits:
\( = \frac{1}{\sqrt{2}} \ln|\sec\left(\frac{\pi}{2}\right) + \tan\left(\frac{\pi}{2}\right)| - \frac{1}{\sqrt{2}} \ln|\sec\left(-\frac{\pi}{2}\right) + \tan\left(-\frac{\pi}{2}\right)| \)
Simplify further:
\( = \frac{1}{\sqrt{2}} \ln|+\infty| - \frac{1}{\sqrt{2}} \ln|-\infty| \)
Since \( \ln|\pm\infty| = +\infty \), the result is:
\( = \frac{1}{\sqrt{2}} \cdot \infty - \frac{1}{\sqrt{2}} \cdot (-\infty) \)
Therefore, the value of the integral is:
\( = \infty \)