What is the integral of 1/sqrt(1 - x^2) dx?

What is the integral of 1/sqrt(1 - x^2) dx?

• A. arccos(x) + C
• B. arctan(x) + C
• C. arcsin(x) + C
• D. sec(x) + C

Answer: (C)

The integral of \( \frac{1}{\sqrt{1 - x^2}} \, dx \) is a standard integral in calculus, often encountered in trigonometric contexts. The correct result of this integral is \( \arcsin(x) + C \). This result can be understood by recognizing that the function \( \frac{1}{\sqrt{1 - x^2}} \) relates to the derivative of the arcsine function. Specifically, if you differentiate \( \arcsin(x) \), you obtain \( \frac{1}{\sqrt{1 - x^2}} \). Therefore, when you perform the integral of \( \frac{1}{\sqrt{1 - x^2}} \), you are essentially reversing this derivative process. Additionally, the arcsine function is defined on the interval [-1, 1], where \( \sqrt{1 - x^2} \) is positive, ensuring that the integral is valid in this domain. This characteristic makes \( \arcsin(x) \) the appropriate choice as it directly corresponds to the antiderivative of the integrand. In the context of the other options, \( \arccos(x) \) is related

The integral of ( \frac{1}{\sqrt{1 - x^2}} , dx ) is a standard integral in calculus, often encountered in trigonometric contexts. The correct result of this integral is ( \arcsin(x) + C ).

This result can be understood by recognizing that the function ( \frac{1}{\sqrt{1 - x^2}} ) relates to the derivative of the arcsine function. Specifically, if you differentiate ( \arcsin(x) ), you obtain ( \frac{1}{\sqrt{1 - x^2}} ). Therefore, when you perform the integral of ( \frac{1}{\sqrt{1 - x^2}} ), you are essentially reversing this derivative process.

Additionally, the arcsine function is defined on the interval [-1, 1], where ( \sqrt{1 - x^2} ) is positive, ensuring that the integral is valid in this domain. This characteristic makes ( \arcsin(x) ) the appropriate choice as it directly corresponds to the antiderivative of the integrand.

In the context of the other options, ( \arccos(x) ) is related