Integration with $\sin()$ and $\cos()$

Question

How do I solve this integration problem?

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

Answer 1

$$ \begin{align} \int\frac{1}{\sin^2 x\cos^2 x}\,dx &= \int\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x}\,dx\\ &= \int\frac{1}{\cos^2 x}\,dx+\int \frac{1}{\sin^2 x}\,dx\\ &= \int \sec^2 x\,dx+\int \csc^2 x\,dx \\ &= \tan x-\cot x+C \end{align} $$

Answer 2

Write $\sin^2 \cos^2 x=\frac14 \sin^2 2x$. Then, we have

$$\int \frac{1}{\sin^2 \cos^2 x}\,dx=4 \int \csc^2 (2x)\,dx=-2\cot (2x)+C$$

Of course, we can use the identity $2\cot 2x =\cot x-\tan x$ to recover the result reported by @juantheron.

Answer 3

For starters, use $\sin2x=2\sin x\cos x$, then let $t=2x$, and finally write $\dfrac1{\sin t}=\dfrac{\sin t}{\sin^2t}=$ $=-\dfrac{\cos't}{1-\cos^2t}$