Question

∫(x²arcsinx)／√(1－x²怎么求)

Answer 1

令y = arcsinx、siny = x、cosy dy = dx
∫ (x²arcsinx)/√(1 - x²) dx
= ∫ [ysin²y/cosy] * [cosy dy]
= ∫ y * (1 - cos2y)/2 dy
= (1/2)∫ y dy - (1/2)∫ ycos2y dy
= y²/4 - (1/4)∫ y d(sin2y)
= y²/4 - (1/4)ysin2y + (1/4)∫ sin2y dy
= y²/4 - (1/4)ysin2y - (1/8)cos2y + C
= (1/4)(arcsinx)² - (1/2)arcsinx * sinycosy - (1/8)(cos²y - sin²y) + C
= (1/4)(arcsinx)² - (1/2)arcsinx * x√(1 - x²) - (1/8)[(1 - x²) - x²] + C
= (1/4)(arcsinx)² - (1/2)x√(1 - x²)arcsinx + x²/4 + C

追问

其实不是为了追问，就想表达一下我的激动之情。你太强了，我一寝室的都不会做 ，就等着赶紧做完了今天交作业呢！感谢感谢…！

追答

加油，你也会做到的