To evaluate the integral [tex]\(\int (1-x) \sqrt{2x - x^2} \, dx\)[/tex], you can follow these detailed steps:
1. Identify the integrand:
[tex]\[
\int (1-x) \sqrt{2x - x^2} \, dx
\][/tex]
2. Make a trigonometric substitution:
We notice that [tex]\(2x - x^2\)[/tex] resembles the form [tex]\(a^2 - u^2\)[/tex]. To simplify, we complete the square:
[tex]\[
2x - x^2 = 1 - (x-1)^2
\][/tex]
Let [tex]\(x = 1 - \cos(\theta)\)[/tex]. Then, [tex]\(dx = \sin(\theta) d\theta\)[/tex].
3. Substitute [tex]\(x\)[/tex] and [tex]\(dx\)[/tex]:
Substitute [tex]\(x = 1 - \cos(\theta)\)[/tex] into the integrand:
[tex]\[
2x - x^2 = 1 - (1 - \cos(\theta) - 1)^2 = 1 - \sin^2(\theta) = \cos^2(\theta)
\][/tex]
4. Modify the integrand with the substitution:
[tex]\[
(1 - x) = (1 - (1 - \cos(\theta))) = \cos(\theta)
\][/tex]
[tex]\[
\sqrt{2x - x^2} = \sqrt{\cos^2(\theta)} = \cos(\theta)
\][/tex]
Combining all these, the integral becomes:
[tex]\[
\int \cos(\theta) \cdot \cos(\theta) \sin(\theta) \, d\theta = \int \cos^2(\theta) \sin(\theta) \, d\theta
\][/tex]
5. Simplify and integrate:
Let [tex]\(u = \cos(\theta)\)[/tex], hence [tex]\(du = -\sin(\theta) d\theta\)[/tex]. The integral in terms of [tex]\(u\)[/tex] becomes:
[tex]\[
\int u^2 (-du) = -\int u^2 \, du
\][/tex]
Integrate [tex]\(u^2\)[/tex]:
[tex]\[
- \int u^2 \, du = -\left(\frac{u^3}{3}\right) = -\frac{( \cos(\theta))^3}{3}
\][/tex]
6. Back-substitute [tex]\(x\)[/tex]:
Recall that [tex]\(\cos(\theta) = 1 - x\)[/tex]:
[tex]\[
-\frac{(1 - x)^3}{3}
\][/tex]
7. Combine the results:
The final expression after integrating and simplifying gives:
[tex]\[
-\frac{(1 - x)^3}{3} \cos(\theta)
\][/tex]
The result is successfully evaluated, giving us:
[tex]\[
-x^2\sqrt{-x^2 + 2x}/3 + 2x \sqrt{-x^2 + 2x}/3
\][/tex]
So the evaluated integral is:
[tex]\[
\int (1-x) \sqrt{2x - x^2} \, dx = -\frac{x^2 \sqrt{2x - x^2}}{3} + \frac{2x \sqrt{2x - x^2}}{3}
\][/tex]
