Salut, ma puteti ajuta la problema 087...?

∫ 1/sinx dx = ∫ sinx / sin²x dx =
= ∫ sinx / (1-cos²x) dx =
= -∫ (cosx)' / (1-cos²x) dx =
= (-1/2)•ln|(cosx-1)/(cosx+1)| =
= (1/2)•ln|(cosx+1)/(cosx-1)|
=> I = (1/2)•ln(1) - (1/2)•ln(3)
=> I = (1/2)•ln(3)