求两道不定积分的解法.1、∫（sinxcosx)^2/((sinx)^3+(cosx)^3)^2 dx2、∫(sinx)

1、
∫（sinxcosx)^2/((sinx)^3+(cosx)^3)^2 dx
=∫（secx)^2(tanx)^2/((tanx)^3+1)^2 dx
=∫(tanx)^2/((tanx)^3+1)^2 dtanx
=1/3∫ 1/((tanx)^3+1)^2 d(tanx)^3
= -1/3[(tanx)^3+1)] + C
2、
(sinx)^2*cosnx
=1/2{（1-cos2x)cosnx}
=1/2{ cosnx - cos2xcosnx }
=1/2{ cosnx - 1/2[cos(n+2)x/2 + cos(n-2)x/2 ] }
=1/2*cosnx - 1/4*cos[(n+2)x/2] - 1/4*cos[(n-2)x/2]
∫(sinx)^2*cosnx dx (n为正整数)
=∫{ 1/2*cosnx - 1/4*cos[(n+2)x/2] - 1/4*cos[(n-2)x/2] } dx
= 1/2n*sinnx - 1/2n*sin[(n+2)x/2] -1/2n*sin[(n-2)x/2] + C