What is the right approach to $\int \frac{1}{\sin^2x\cos^2x} dx$?

Question

This is a problem related to trigonometry and indefinite integration.what is the right approach to this problem.. What is the basic idea?

$$\int \frac{1}{\sin^2x\cos^2x} dx$$

Answer 1

HINT:

$$\dfrac1{\sin^2x\cos^2x}=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=?$$

Alternatively,

$$\dfrac1{\sin^2x\cos^2x}=\dfrac4{\sin^22x}=4\csc^22x$$

Can you explain why the results are seemingly different?

Answer 2

$$\int \frac{dx}{\sin^2 x \cos^2 x}$$ $$=\int \frac{4dx}{4\sin^2 x \cos^2 x}$$ $$=\int \frac{4dx}{\sin^2 2x }$$ $$=\int 4 cosec^2 2x dx$$ $$=-4\frac{\cot 2x}{2} + c$$ $$=-2 \cot 2x + c$$ Where c is a constant of integration

This is just one of the ways. Lab's is another way.