What is the integral of arccos(x)?

What is the integral of arccos(x)?

• A. x*arccos(x) + sqrt(1-x^2) + C
• B. x*arccos(x) - sqrt(1-x^2) + C
• C. ln(1 + x) + x*arccos(x) + C
• D. arccos(x) + C

Answer: (B)

To find the integral of arccos(x), it is useful to apply integration by parts, where we let one part be arccos(x) and the other part be the differential of x. Using integration by parts, we set: - u = arccos(x), which implies that du = -1/sqrt(1 - x^2) dx. - dv = dx, which gives v = x. Now applying integration by parts: ∫u dv = u*v - ∫v du. Substituting in the values, we have: ∫arccos(x) dx = x*arccos(x) - ∫x * (-1/sqrt(1-x^2)) dx. The remaining integral, ∫x * (1/sqrt(1-x^2)) dx, can be approached by substituting: Let w = 1 - x^2, then dw = -2x dx, or dx = -dw/(2x). This integral simplifies to: ∫x/sqrt(1-x^2) dx = -∫(sqrt(1-w)/2) dw = -sqrt(1-x^2). Putting it all back together, we get

To find the integral of arccos(x), it is useful to apply integration by parts, where we let one part be arccos(x) and the other part be the differential of x.

Using integration by parts, we set:

• u = arccos(x), which implies that du = -1/sqrt(1 - x^2) dx.

• dv = dx, which gives v = x.

Now applying integration by parts:

∫u dv = u*v - ∫v du.

Substituting in the values, we have:

∫arccos(x) dx = x*arccos(x) - ∫x * (-1/sqrt(1-x^2)) dx.

The remaining integral, ∫x * (1/sqrt(1-x^2)) dx, can be approached by substituting:

Let w = 1 - x^2, then dw = -2x dx, or dx = -dw/(2x).

This integral simplifies to:

∫x/sqrt(1-x^2) dx = -∫(sqrt(1-w)/2) dw = -sqrt(1-x^2).

Putting it all back together, we get