Question

能给我答案吗 ∫x³arccosx/(√1-x²)dx

Answer 1

求不定积分∫[(x³arccosx)/(√1-x²)]dx
解：令arccosx=u，则x=cosu，dx=-sinudu；代入原式得：
原式=-∫[ucos³usinu/√(1-cos²u)]du=-∫ucos³udu
=-(1/4)∫u(cos3u+3cosu)du=-(1/4)[∫ucos3udu+3∫ucosudu]
=-(1/4)[(1/3)∫ud(sin3u)+3∫udsinu]=-(1/4)[(1/3)usin3u-(1/3)∫sin3udu+3usinu-3∫sinudu]
=-(1/12)usin3u+(1/36)∫sin3ud(3u)-(3/4)usinu-(3/4)cosu
=-(1/12)usin3u-(1/36)cos3u-(3/4)usinu-(3/4)cosu+C
=-(1/12)(arccosx)sin(3arccosx)-(1/36)cos(3arccosx)-(3/4)(arccosx)√(1-x²)-(3/4)x+C
=-(1/12)(arccosx)[3√(1-x²)-4√(1-x²)³]-(1/36)(4x³-3x)-(3/4)(arccosx)√(1-x²)-(3/4)x+C
=-(1/4)(arccosx)√(1-x²)+(1/3)(arccosx)√(1-x²)³]-(1/36)(4x³-3x)-(3/4)(arccosx)√(1-x²)-(3/4)x+C
=-(arccosx)√(1-x²)+(1/3)(arccosx)√(1-x²)³]-(1/36)(4x³-3x)-(3/4)x+C
【运算中使用了以下一些公式：
sin3x=3sinx-4sin³x；cos3x=4cos³x-3cosx；cos³x=(1/4)(cos3x+3cosx)；
sin(arccosx)=√(1-x²)；cos(arccosx)=x；】