Question

Evaluate the integral: \[ \int e^{\cos^{-1}x} \left[ x - \sqrt{1 - x^2} \right] \frac{dx}{\sqrt{1 - x^2}}. \]

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

Hint:

For integrals involving inverse trigonometric functions, use substitution to simplify the integrand.

Answer 1

The Correct Option is B

Step 1: Simplify the integrand.
We are given the integral: \[ I = \int e^{\cos^{-1} x} \left[ x - \sqrt{1 - x^2} \right] \frac{dx}{\sqrt{1 - x^2}}. \] We can rewrite the expression \( \left[ x - \sqrt{1 - x^2} \right] \) as \( - \sqrt{1 - x^2} \), and then simplify the integral to: \[ I = \int -x e^{\cos^{-1} x} \, dx. \]
Step 2: Use substitution.
We know that \( \cos^{-1} x \) is the inverse of \( \cos x \), so we can make a substitution: \[ u = \cos^{-1} x \quad \Rightarrow \quad du = \frac{-dx}{\sqrt{1 - x^2}}. \] Substituting into the integral, we get: \[ I = -x e^{u} + c. \]
Step 3: Conclusion.
Thus, the solution is \( -x e^{\cos^{-1} x} + c \), which corresponds to option (B).