Question

∫sinx/cosx√(5-4cosx)dx

Answer 1

设√(5-4cosx)=t,则sinxdx=tdt/2

∴原式=∫(tdt/2)/[t(5-t²)/4]

=2∫dt/(5-t²)

=(1/√5)∫[1/(√5+t)+1/(√5-t)]dt

=(1/√5)[ln│√5+t│-ln│√5-t│]+C (C是积分常数)

=(1/√5)ln│(√5+t)/(√5-t)│+C

=(1/√5)ln│[√5+√(5-4cosx)]/[√5-√(5-4cosx)]│+C