∫sin^4(t)(1-sin^2(t))dt ( 范围是0到π/2)

∫(sint)^4*(1-sin²t)dt =∫(sin²t)²*cos²tdt
=∫[(1-cos(2t))/2]²*[(1+cos(2t))/2]dt (应用倍角公式)
=(1/8)∫[1-cos(2t)]*[1-cos²(2t)]dt (应用两数平方差公式)
=(1/8)∫[1-cos(2t)]*sin²(2t)dt
=(1/8)∫[sin²(2t)-cos(2t)*sin²(2t)]dt
=(1/8)∫[(1-cos(4t))/2-cos(2t)*sin²(2t)]dt
=(1/8){(1/2)∫[1-cos(4t)]dt-∫[cos(2t)*sin²(2t)]dt}
=(1/16){∫[1-cos(4t)]dt-∫[sin²(2t)]d[sin(2t)]}
=(1/16)[t-sin(4t)/4-sin³(2t)/3]│
=(1/16)(π/2-0-0-0+0+0)
=π/32.