∫ ( 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}
OR
∫ 1 + t a n 2 x d x = ∫ 1 + s i n 2 x c o s 2 x d x = ∫ c o s 2 x + s i n 2 x c o s 2 x d x cos 2 x + s i n 2 x = 1 ∫ 1 c o s 2 x d x = ∫ sec 2 x d x = t a n x + C {\displaystyle \int _{}^{}1+tan^{2}xdx=\int _{}^{}1+{\frac {sin^{2}x}{cos^{2}x}}dx=\int _{}^{}{\frac {cos^{2}x+sin^{2}x}{cos^{2}x}}dx\cos ^{2}x+sin^{2}x=1\int _{}^{}{\frac {1}{cos^{2}x}}dx=\int _{}^{}\sec ^{2}xdx=tanx+C}
