设3t=sinx,dt = (1/3)cosx dx ,被积函数 = [(1/9)(sinx)^2]/cosx∴积分 = (1/9)∫ [(1/cosx)-cosx]dx,
而∫(1/cosx) dx = ∫d(sinx)/[1-(sinx)^2]
= (1/2)∫[1/(1 -sinx) + 1/(1 +sinx)]d(sinx)
=(1/2)ln[(1+ sinx)/(1- sinx)]
∴原积分 = (1/9){(1/2)ln[(1+ sinx)/(1- sinx)] - sinx} + C
= (1/9){(1/2)ln[(1+3t)/(1-3t)] - 3t} + C