Question

integration of (2 - 3sinx)/cos²x . dx

Answer 1

you can see your answer

Answer 2

given,

$\bf{\int{\frac{(2-3sinx)}{cos^2x}}\,dx}\\\\=\bf{\int{\frac{2}{cos^2x}}\,dx-\int{\frac{3sinx}{cos^2x}}\,dx}\\\\=\bf{2\int{sec^2x}\,dx-3\int{\frac{sinx}{cosx}.\frac{1}{cosx}}\,dx}\\\\=\bf{2\int{sec^2x}\,dx-3\int{tanx.secx}\,dx}\\\\=\bf{2tanx-3secx+C}$

hence, answer is 2tanx - 3secx + C