原式=∫1/(cosx)^2dx/[1+3/(cosx)^2]
=∫(secx)^2dx/[3(secx)^2+1]
=∫d(tanx)/[3(tanx)^2+4]
设tanx=u,
原式=∫du/(4+3u^2)
=(1/4)*(2/√3)∫d(√3u/2)/[1+(√3u/2)^2]
=1/(2√3)arctan(√3u/2)+C
=(√3/6)arctan(√3tanx/2)+C
原式=∫1/(cosx)^2dx/[1+3/(cosx)^2]
=∫(secx)^2dx/[3(secx)^2+1]
=∫d(tanx)/[3(tanx)^2+4]
设tanx=u,
原式=∫du/(4+3u^2)
=(1/4)*(2/√3)∫d(√3u/2)/[1+(√3u/2)^2]
=1/(2√3)arctan(√3u/2)+C
=(√3/6)arctan(√3tanx/2)+C