求不定积分∫sin²xdx
原式=∫[(1-cos2x)/2]dx=(1/2)x-(1/2)∫cos2xdx=(1/2)x-(1/4)∫cos2xd(2x)=(1/2)x-(1/4)sin2x+C
关于∫sinⁿxdx有递推公式：
∫sinⁿxdx=-(sinⁿֿ¹xcosx)/n+[(n-1)/n]∫sinⁿֿ²xdx.
∫sin⁴xdx=-∫sin³xd(cosx)=-[sin³xcosx-3∫cos²xsin²xdx]=-sin³xcosx+3[∫(1-sin²x)sin²xdx]
=-sin³xcosx+3∫sin²xdx-3∫sin⁴xdx
故移项有4∫sin⁴xdx=-sin³xcosx+3∫sin²xdx=-sin³xcosx+3[(1/2)x-(1/4)sin2x]+C
=-sin³xcosx-(3/4)sin2x+(3/2)x+C=-sin³xcosx-(3/2)sinxcosx+(3/2)x+C
=-sinxcosx(sin²x+3/2)+(3/2)x+C₁
故∫sin⁴xdx=-(1/4)sinxcosx(sin²x+3/2)+6x+4C₁=-(1/8)sin2x(sin²x+3/2)+6x+C.