∫(π⼀2-0) 1⼀(sinx+cosx)

2026年09月27日 04:29
有1个网友回答
网友(1):

∫(0->π/2) dx/(sinx+cosx)
=(1/√2)∫(0->π/2) dx/sin(x+π/4)
=(1/√2)∫(0->π/2) csc(x+π/4) dx
=(1/√2) ln|csc(x+π/4) - cot(x+π/4)| | (0->π/2)
=(1/√2) [ ln|√2 +1| - ln|√2-1| ]
=(1/√2)ln(3+2√2)