I got $3$ different answers by using $3$ different methods while solving, $$\int \frac{1}{\sin^2x \cos^2x}\ dx$$
Just want to know if there is any problem in any of the method.
Method 1:
$$\int \frac{1}{\sin^2x \cos^2x}\ dx\\ =\int \frac{\sin^2x + \cos^2x}{\sin^2x \cos^2x}\ dx\\ = \int \sec^2x + \csc^2x\ dx \\= \tan x - \cot x + C $$
Method 2:
$$\int \dfrac{1}{\sin^2x \cos^2x}\ dx\\=\int \dfrac{\sec^4x}{\tan^2x}\ dx \\ = \int \dfrac{(\tan^2x + 1)\sec^2x}{\tan^2x}\ dx\\ =\int \dfrac{(t^2 + 1)}{t^2}\ dt\quad {\text {where }}t = \tan(x)\\= \tan(x) - \frac{1}{\tan(x)} + C $$