求不定积分∫x^2arcsinx/√(1-x^2)

题目

求不定积分∫x^2arcsinx/√(1-x^2)

答案

换元,t = arcsinx, dx = cost dt
I = ∫ t sin²t dt = (1/2) ∫ t (1﹣cos2t) dt
= (1/4) t² ﹣(t/4)sin2t + (1/4) ∫ sin2t dt
= (1/4) t² ﹣(t/4)sin2t ﹣ (1/8) cos2t + C
= (1/4)arcsin²x ﹣(1/2) x √(1-x²) arcsinx ﹣ (1/8) (1﹣2x²) + C
= (1/4)arcsin²x ﹣(1/2) x √(1-x²) arcsinx + (1/4) x² + C