求定积分dx/(1+(cosx)^2) pai/2

网友回答

∫(0->π/2) dx/(1+cos²x)
= ∫(0->π/2) dx/(sin²x+cos²x+cos²x)
= ∫(0->π/2) dx/(sin²x+2cos²x)
= ∫(0->π/2) sec²x/(tan²x+2)
= ∫(0->π/2) d(tanx)/(2+tan²x)
= (1/√2)arctan(tanx/√2),(0->π/2)= (1/√2)arctan[(1/√2)tan(π/2)] - 0
= (1/√2)arctan(∞)
= (1/√2)(π/2)
= π/(2√2)