∫cos^2√xdx
=∫(cos2√x+1)/2 dx
=1/2∫cos2√xdx+1/2∫dx
=1/2∫cos2√xdx+x/2
单独求
∫cos2√xdx
令√x=t
x=t^2
dx=2tdt
则原代化为
∫cos2t*2tdt
=∫tdsin2t
=tsin2t-∫sin2tdt
=tsin2t+cos2t+C
=√xsin2√x+cos2√x+C
所以原式=1/2*(√xsin2√x+cos2√x+x)+C