Integral of 1/root(2x-x^2)dx
2x-x^2 = 1 - (1-x)^2
let u = 1-x
du = -dx
now you have
∫-du/√(1-u^2) = -arcsin(u) = -arcsin(1-x)+C
*This is not an answer*
Thanks Steve that really helps!!!
*This is not an answer*
To find the integral of the function 1/√(2x - x^2) dx, we can use the method of partial fractions.
Step 1: Factorize the denominator.
The denominator √(2x - x^2) can be written as √(x(2 - x)) or √(x(x - 2)).
Step 2: Decompose the fraction into partial fractions.
Since the degree of the denominator is 2, we can decompose the fraction into two partial fractions. Let's assume the partial fractions are A/√x and B/√(x - 2).
1/√(x(x - 2)) = A/√x + B/√(x - 2)
Step 3: Clear the denominators.
To clear the denominators, we need to multiply both sides of the equation by the least common multiple (LCM) of the denominators. In this case, the LCM is √x * √(x - 2).
√x * √(x - 2) * (1/√(x(x - 2))) = √x * √(x - 2) * (A/√x + B/√(x - 2))
Simplifying the left side and canceling out the square root terms, we get:
1/√(x - 2) - 1/√x = A * √(x - 2) + B * √x
Step 4: Solve for A and B.
To find the values of A and B, we can equate the coefficients of the square root terms on both sides of the equation.
1) Coefficient of √(x - 2): -1/√(x - 2) = A
2) Coefficient of √x: -1/√x = B
From equation 1, we get:
A = -1/√(x - 2)
From equation 2, we get:
B = -1/√x
Step 5: Rewrite the integral using the partial fraction decomposition.
Now that we have the values of A and B, we can rewrite the integral:
∫(1/√(2x - x^2)) dx = ∫(-1/√(x - 2)) dx + ∫(-1/√x) dx
Step 6: Evaluate the integrals.
Integrating each term separately, we get:
∫(-1/√(x - 2)) dx = -2√(x - 2)
∫(-1/√x) dx = -2√x
Therefore, the integral of 1/√(2x - x^2) dx is:
∫(1/√(2x - x^2)) dx = -2√(x - 2) - 2√x + C
where C is the constant of integration.