∫sin⁻¹x dx equals:
∫sin⁻¹x dx equals:
- A. x sin⁻¹x + √(1−x²) + C
- B. x sin⁻¹x − √(1−x²) + C
- C. x sin⁻¹x + C
- D. cos⁻¹x + C
Answer: A) x sin⁻¹x + √(1−x²) + C
Explanation: Using integration by parts: u = sin⁻¹x, dv = dx. du = dx/√(1−x²), v = x. ∫ = x sin⁻¹x − ∫x/√(1−x²) dx. Substituting t = 1−x² gives +√(1−x²) + C.
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