∫ ( 1 + tan 2 α ) d α = ∫ sec 2 α d α = tan α + C {\displaystyle \int (1+\tan ^{2}{\alpha })\,d\alpha =\int \sec ^{2}\alpha \,d\alpha =\tan \alpha +C}
Note: 1 + tan 2 α = sec 2 α {\displaystyle 1+\tan ^{2}{\alpha }=\sec ^{2}\alpha }
Or,
∫ ( 1 + tan 2 α ) d α = ∫ ( 1 + s i n 2 α c o s 2 α ) d α = ∫ c o s 2 α + s i n 2 α c o s 2 α d α cos 2 x + s i n 2 x = 1 ∫ 1 c o s 2 x d x = ∫ sec 2 x d x = tan x + C {\displaystyle \int (1+\tan ^{2}{\alpha })\,d\alpha =\int \left(1+{\frac {sin^{2}\alpha }{cos^{2}\alpha }}\right)d\alpha =\int {\frac {cos^{2}\alpha +sin^{2}\alpha }{cos^{2}\alpha }}d\alpha \cos ^{2}x+sin^{2}x=1\int {\frac {1}{cos^{2}x}}dx=\int \sec ^{2}xdx=\tan {x}+C}
