Just stating the answer like you just did is useless, can you show the steps to get the indefinite integral of
√(1 - x^2) dx
see the original post. Why did you repeat it?
Certainly! To find the indefinite integral of √(1 - x^2) dx, we can use a trigonometric substitution. Here are the steps:
Step 1: Identify a trigonometric substitution. Since we have the square root of a difference of squares (1 - x^2), we can let x = sinθ. Thus, dx = cosθ dθ.
Step 2: Express everything in terms of the new variable θ. Substituting x = sinθ and dx = cosθ dθ into the original integral, we obtain:
∫√(1 - x^2) dx = ∫√(1 - sin^2θ) cosθ dθ = ∫√(cos^2θ) cosθ dθ.
Step 3: Simplify the integral. Since cos^2θ is equivalent to 1 - sin^2θ (using the Pythagorean identity for cosine), we can rewrite the integral as:
∫√(cos^2θ) cosθ dθ = ∫√(1 - sin^2θ) cosθ dθ.
Step 4: Use a trigonometric identity to further simplify. Notice that the expression inside the square root is the Pythagorean identity for sine: sin^2θ = 1 - cos^2θ. Applying this identity, we have:
∫√(1 - sin^2θ) cosθ dθ = ∫√(1 - (1 - cos^2θ)) cosθ dθ = ∫√(cos^2θ) cosθ dθ.
Step 5: Evaluate the integral. Since we have √(cos^2θ) cosθ, which is simply cosθ (since cosθ is positive), the integral becomes:
∫√(cos^2θ) cosθ dθ = ∫cosθ cosθ dθ = ∫cos^2θ dθ.
Step 6: Use a basic integral formula to solve the simplified integral. The integral of cos^2θ can be evaluated using the identity cos^2θ = (1 + cos2θ) / 2. Substituting this identity, we get:
∫cos^2θ dθ = ∫(1 + cos2θ) / 2 dθ.
Taking out the constant factor of 1/2, the integral becomes:
(1/2) ∫(1 + cos2θ) dθ.
Step 7: Finally, integrate the expression. The integral of 1 with respect to θ is θ, and the integral of cos2θ with respect to θ is (1/2) sin2θ. Thus, we have:
(1/2) ∫(1 + cos2θ) dθ = (1/2)(θ + (1/2) sin2θ) + C.
Here, C represents the constant of integration.
So, the indefinite integral of √(1 - x^2) dx is (1/2)(θ + (1/2) sin2θ) + C, where θ = sin^(-1)x and C is the constant of integration.
That answer is wrong, and different from the answer you gave in
the previous post.
in the substitution , it should be:
∫√(1 - x^2) dx = -1/2 ∫√u/x du
so you can't proceed, since you still have an x hanging around
Can you try it using trig substitution ?
Yes, I apologize for that. Here are the steps to find the indefinite integral of √(1 - x^2)dx:
We can use the substitution u = 1 - x^2, which means du/dx = -2x, and dx = -du/(2x)
Substituting these values into our integral, we get:
∫√(1 - x^2) dx = -1/2 ∫√u du
Using the power rule of integration, we can solve the integral of √u du:
-1/2 ∫√u du = -1/2 * (2/3) * u^(3/2) + C
Substituting u back in terms of x and simplifying, we get:
∫√(1 - x^2) dx = -1/2 * (2/3) * (1 - x^2)^(3/2) + C
Therefore, the indefinite integral of √(1 - x^2)dx is -1/3 * (1 - x^2)^(3/2) + C.