Question

请教x∧2*(sinx)∧3的原函数

Answer 1

原理:积分:u'v du=uv-积分:uv' dv
积分:x^2*(sinx)^3 dx
=积分:x^2*(sinx)*[1-(cosx)^2] dx
=积分:x^2*sinx dx + 积分:x^2*(sinx)*(cosx)^2 dx
=(-cosx)*x^2+积分:2x(cosx) dx +(1/3)(x^2)(cosx)^3-积分:(2/3)*x*(cosx)^3 dx
=(-cosx)*(x^2)+(2x)(sinx)-积分:2sinx dx + (1/3)(x^2)(cosx)^3-积分:(2/3)*x*(cosx)*[1-(sinx)^2] dx
=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-积分:(2/3)*x*(cosx) dx + 积分:(2/3)*x*(cosx)*(sinx)^2 dx
=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-(2/3)*x*(sinx)+积分:(2/3)(sinx) dx + (2/9)*x*(sinx)^3 - 积分:(2/9)*(sinx)^3 dx
=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-(2/3)*x*(sinx)-(2/3)(cosx) + (2/9)*x*(sinx)^3 -积分:(2/9)*(sinx)[1-(cosx)^2] dx
=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-(2/3)*x*(sinx)-(2/3)(cosx) + (2/9)*x*(sinx)^3 -积分:(2/9)(sinx) dx+积分:(2/9)(sinx)(cosx)^2 dx
=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-(2/3)*x*(sinx)-(2/3)(cosx) + (2/9)*x*(sinx)^3 +(2/9)(cosx)-(2/27)(cosx)^3
=-(x^2)cosx+(4/3)xsinx+(14/9)cosx+(1/3)(x^2)(cosx)^3+(2/9)x(sinx)^3-(2/27)(cosx)^3+C