Question

integration of [sec²x]/[cosec²x] . dx

Answer 1

you can see your answer

Answer 2

$\bf{I=\int{\frac{sec^2x}{cosec^2}}\,dx}\\\\=\bf{\int{\frac{\frac{1}{cos^2x}}{\frac{1}{sin^2}}}\,dx}\\\\=\bf{\int{\frac{sin^2x}{cos^2}}\,dx}\\\\=\bf{\int{tan^2x}\,dx}\\\\=\bf{\int{(sec^2x-1)}\,dx}\\\\=\bf{\int{sec^2x}\,dx-\int{dx}}\\\\=\bf{tanx-x+C}$

hence, I = tanx - x + C