tìm tích phân (2cosx-1/cos^2x)dx
Giải thích
\(\int {\left( {2\cos x - \frac{1}{{{{\cos }^2}x}}} \right)} dx = 2\int {\cos } xdx - \int {\frac{1}{{{{\cos }^2}x}}} dx = 2\sin x - \tan x + C\).
\(\int {\left( {2\cos x - \frac{1}{{{{\cos }^2}x}}} \right)} dx = 2\int {\cos } xdx - \int {\frac{1}{{{{\cos }^2}x}}} dx = 2\sin x - \tan x + C\).