cos²√xdx求不定积分

Question

如题所述

Answer 1

设√x=t，则x=t²，dx=2tdt
原式=∫cos²t*2tdt=∫(2cos²t)*tdt
=∫(cos2t+1)*tdt
=∫cos2t*tdt+∫tdt
=(1/2)∫td(sin2t)+(1/2)t²
=(1/2)[t*sin2t-∫sin2tdt]+(1/2)t²
=(1/2)tsin2t-(1/2)∫sin(2t)dt+(1/2)t²
=(1/2)tsin2t+(1/4)cos(2t)+(1/2)t²+C
=(1/2)√x*sin(2√x)+(1/4)cos(2√x)+(1/2)x+C本回答被提问者采纳