tìm tích phân tan bình x dx
Giải thích
\(\int {{{\tan }^2}} x\;{\rm{d}}x = \int {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)} {\rm{d}}x = \int {\frac{1}{{{{\cos }^2}x}}} \;{\rm{d}}x - \int {\rm{d}} x = \tan x - x + C\)
\(\int {{{\tan }^2}} x\;{\rm{d}}x = \int {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)} {\rm{d}}x = \int {\frac{1}{{{{\cos }^2}x}}} \;{\rm{d}}x - \int {\rm{d}} x = \tan x - x + C\)