Question

If d/dx(Arccos(x)) gives -1/√(1 - x^2), what is the result of ∫(-1/√(1 - x^2))dx?

A) Arcsin(x) + C
B) Arccos(x) + C
C) −Arcsin(x) + k
D) −Arccos(x) + k

Answer 1

The result of the integral of -1/√(1 - x^2) is -arcsin(x) + C.

Step-by-step explanation:

To find the integral of -1/√(1 - x^2) with respect to x, we can use the substitution method. Let's substitute u = arccos(x). Then, we can find du = -1/√(1 - x^2) dx. Rearranging the equation gives us dx = -√(1 - x^2) du. Substituting these values into the integral, we have:

∫(-1/√(1 - x^2))dx = ∫-√(1 - x^2) du

Now we can integrate this expression, which gives us:

∫-√(1 - x^2) du = -∫√(1 - u^2) du = -arcsin(u) + C = -arcsin(arccos(x)) + C =

-arcsin(x) + C