(1-sinx平方)除以sinx从6分之π到2分之π的定积分

网友回答

∫(π/6→π/2) (1 - sin²x)/sinx dx
= ∫(π/6→π/2) (cscx - sinx) dx
= ln(cscx - cotx) + cosx |(π/6→π/2)
= 0 - [ln(2 - √3) + √3/2]
= - ln(2 - √3) - √3/2