求不定积分∫cos根号下x+1dx
人气:110 ℃ 时间:2019-12-06 20:11:14
解答
∫cos√(x+1)d(x+1)
=∫cos√(x+1)d(√(x+1)^2
√(x+1)=t
=∫costdt^2=∫2tcostdt=∫2tdsint=2tsint-2∫sintdt=2tsint+2cost+C
=2√(x+1)sin√(x+1)+2cos√(x+1) +C
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