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
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