用换元法计算不定积分∫cos(3x+2)dx
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用换元法计算不定积分∫cos(3x+2)dx
![用换元法计算不定积分∫cos(3x+2)dx](/uploads/image/z/3778819-43-9.jpg?t=%E7%94%A8%E6%8D%A2%E5%85%83%E6%B3%95%E8%AE%A1%E7%AE%97%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%ABcos%283x%2B2%29dx)
令t=3x+2,则dt=3dx→dx=1/3·dt
∫cos(3x+2)dx
=∫cost·1/3·dt
=1/3·∫cost dt
=1/3·sint+C
=1/3·sin(3x+2)+C
再问: 则dt=3dx→dx=1/3·dt???不懂
再答: t=3x+2,则dt=d(3x+2)=3dx
∫cos(3x+2)dx
=∫cost·1/3·dt
=1/3·∫cost dt
=1/3·sint+C
=1/3·sin(3x+2)+C
再问: 则dt=3dx→dx=1/3·dt???不懂
再答: t=3x+2,则dt=d(3x+2)=3dx