不定积分∫1+cos2t⼀2dt如何得到t⼀2+sin2t⼀4+C的求详细过程

2025-01-05 17:06:56
推荐回答(2个)
回答1:

∫ (1 + cos2t)/2 dt
= (1/2)∫ dt + (1/2)∫ cos2t dt,分配律
= (1/2) * t + (1/2)∫ cos(2t) (1/2)d(2t)
= t/2 + (1/4)∫ cos(u) du,u = 2t
= t/2 + (1/4)sin(u) + C,cosx的积分是sinx + C
= t/2 + (1/4)sin(2t) + C,将u = 2t代回,其实熟练了这步可以不要的。

回答2:

∫1+cos2t/2dt=1/4∫2+cos2td2t=1/4(2t+sin2t)+C=t/2+sin2t/4+C