Question

Evaluate: \(\displaystyle \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right) dx = \)

• A. \( e^x \frac{1}{\sqrt{1 - x^2}} + c \)
• B. \( e^x \sin^{-1} x + c \)
• C. \( 2 e^x + c \)
• D. \( e^x \cos^{-1} x + c \)

Hint:

Look for expressions matching the derivative of a product. The product rule helps simplify integrals like: \[ \int f(x) g(x) + f(x) g'(x) \, dx = f(x) g(x) + c \]

Answer 1

The Correct Option is B

Step 1: Observe the integrand: \[ e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right) \] Step 2: Note that the derivative of \( \sin^{-1} x \) is: \[ \frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}} \] Step 3: Recognize the integrand as: \[ e^x \sin^{-1} x + e^x \frac{d}{dx}(\sin^{-1} x) \] Step 4: This suggests the integrand is the derivative of the product \( e^x \sin^{-1} x \) by the product rule: \[ \frac{d}{dx} \left( e^x \sin^{-1} x \right) = e^x \sin^{-1} x + e^x \frac{1}{\sqrt{1 - x^2}} \] Step 5: Therefore, \[ \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right) dx = e^x \sin^{-1} x + c \] Final Answer: \( \boxed{e^x \sin^{-1} x + c} \)