∫sin√道tdt/√t =2∫sin√td(√t) =-2cos√t+C ∫xdx/√(2-3x²) =-⅙∫d(2-3x²)/√(2-3x²) =-⅓√(2-3x²)+C ∫cos²(ω专t+φ属)sin(ωt+φ)dt =(-1/ω)∫cos²(ωt+φ)d[cos(ωt+φ)] =-cos³(ωt+φ)/(3ω)+C