求In=∫(secx)^ndx的递推公式

Question

求In=∫(secx)^ndx的递推公式

Answer 1

用分部积分法,
令u=1/sinx,dv=sinx/(cosx)^n
则du=-cosx/(sinx)^2,v=1/[(n-1)(cosx)^(n-1)]
则
In
=∫udv
=uv-∫vdu
=uv+1/(n-1)∫1/[(cosx)^(n-2)(sinx)^2]
=uv+1/(n-1)∫1/{(cosx)^(n-2)[1-(cosx)^2]}
=uv+1/(n-1)∫1/[(cosx)^(n-2)]-1/(n-1)∫1/[(cosx)^n]
=uv+I(n-2)/(n-1)-In/(n-1)