Question

算不定积分∫1/(4(sinx)^2+(cosx)^2) dx

Answer 1

∫1/(4(sinx)^2+(cosx)^2) dx=∫1/(3(sinx)^2+1) dx=∫2/(5-3cos2x) dx

对于1/(a+bcosx)的积分有公式

原函数为2/(a+b)[(a+b)/(a-b)]^1/2arctan【{(a-b)/(a+b)}^1/2tanx/2】+C