Question

求sinx*sin2x*sin3xdx的不定积分

Answer 1

∫sinx·sin2x·sin3xdx
=∫sinx·sin2x·sin（x+2x）dx
=sinx·sin2x·（sinxcos2x+sin2xcosx）dx
=∫sinx·sin2x·（sinx(cos^2x-sin^2x)+2sinx cosx cosx）dx
=∫sinx 2sinx cosx[sinxcos^2x-sinx^3x+2 sinx cos^2x ]dx
=∫2sin^2x(sinxcos^2x-sinx^3x+2 sinx cos^2x )d(sinx)
=∫2sin^2x[sinx(1-sin^2x)-sinx^3x+2 sinx (1-sin^2x)]d(sinx)
=∫2sin^2x(3sinx-4sinx^3x)d(sinx)
=∫（6sin^3x-8sin^5x）dx
=3/2(sinx)^4-4/3(sinx)^6+C