Question

Evaluate the integral: \[ \int \frac{\sin^1 x}{\sqrt{1 - x^2}} \, dx \]

• A. \( \frac{1}{2} \)
• B. \( \frac{\sin^2 x}{2} \)
• C. \( \frac{\cos^2 x}{2} \)
• D. \( \frac{\sin x}{2} \)

Hint:

Recognize the standard trigonometric integrals like \( \int \frac{\sin x}{\sqrt{1 - x^2}} \, dx \), which directly lead to the square of the sine function over 2.

Answer 1

The Correct Option is B

We are asked to evaluate the integral: \[ I = \int \frac{\sin^1 x}{\sqrt{1 - x^2}} \, dx \]

1. Step 1: Recognize the form of the integrand The given integral is closely related to the standard integral form: \[ \int \frac{\sin x}{\sqrt{1 - x^2}} \, dx \] This is a standard trigonometric integral. We know from calculus that: \[ \int \frac{\sin x}{\sqrt{1 - x^2}} \, dx = \frac{\sin^2 x}{2} \]

2. Step 2: Apply the formula Using the known formula, the result of the integral is: \[ \frac{\sin^2 x}{2} \] Thus, the answer is: \[ I = \frac{\sin^2 x}{2} \]