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.
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
3 answers
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 ?
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 ?
see the original post. Why did you repeat it?