求不定积分dx/(5+4sinx)cosx

Question

求不定积分dx/(5+4sinx)cosx

Answer 1

原式=∫dx/(5+4sinx)cosx
=∫cosxdx/[(5+4sinx)(cosx)^2]
=∫d(sint)/[(5+4sinx)(1-(sinx)^2)]
=∫du/[(5+4u)(1-u^2)],u=sinx
=∫du[a/(5+4u)+b/(1+u)+c/(1-u)]
求a,b,c:
a(1-u^2)+b(5+4u)(1-u)+c(5+4u)(1+u)=1
令u=1,得：c*9*2=1,得c=1/18
令u=-1,得：b*1*2=1,得：b=1/2
令u=-5/4,得：a(1-25/16)=1,得：a=-16/9
因此原式=a/4ln|u+5/4|+bln|1+u|-cln|1-u|+C
=-4/9ln|sinx+5/4|+1/2ln|1+sinx|-1/18ln|1-sinx|+C