求∫arcsinx/[(1-x^2)]^1/2*x^2 dx

1个回答

  • 令y = arcsinx、siny = x、dx = cosy dy

    ∫ (x²arcsinx)/√(1 - x²) dx

    = ∫ ysin²y/(cosy) * (cosy dy)

    = ∫ y * (1 - cos2y)/2 dy

    = (1/2)∫ y dy - (1/2)∫ ycos2y dy

    = y²/4 - (1/2)(1/2)∫ y d(sin2y)

    = y²/4 - (1/4)ysin2y + (1/4)∫ sin2y dy

    = y²/4 - (1/4)ysin2y - (1/8)cos2y + C

    = y²/4 - (y/2)sinycosy - (1/8)(cos²y - sin²y) + C

    = (1/4)(arcsinx)² - (x/2)√(1 - x²)arcsinx - (1/8)[(1 - x²) - x²] + C

    = (1/4)(arcsinx)² - (x/2)√(1 - x²)arcsinx - (1/8)(1 - 2x²) + C