# Integral of $\frac{1}{\sin^2(x)\cos^2(x)}$

$$I = \int\frac{1}{\sin^2(x)\cos^2(x)}$$ I have tried the following:
$$\sin^2x = \sin x \cdot \sin x = \frac{1}{2}(1 - \cos2x)$$ $$\cos^2x = \cos x \cdot \cos x = \frac{1}{2}(1 + \cos2x)$$
The integral becomes:
$$I = 4\int\frac{1}{1 - \cos^2(2x)} = 4\int\frac{1}{\sin^2(2x)}$$
Substitute $$u = 2x \implies du = 2dx$$
$$I = 2\int\frac{1}{\sin^2u} = -2\cot(u)$$
Plugging back x I get:
$$I = -2\cot(2x)$$
I tried plugging in the integral into an integral-calculator and the answer was: $$\tan(x) - \cot(x)$$. Can you help me identify what I did wrong?

• I tried plugging upper and lower bound to the integral and it turns out the answer is the same! $-2cot(2x)$ = $tanx- cotx$ – Radu Gabriel Aug 30 '19 at 11:57
• Please use MathJax for trigonometric functions like \cos – sera Aug 30 '19 at 12:02
• I will, I did not know about it, sorry! – Radu Gabriel Aug 30 '19 at 12:10

Your result is correct. Here's another approach:

$$\sin2x=2\sin x\cos x\implies \sin^2x\cos^2x=\frac14\sin^22x$$

and since $$\;(\cot x)'=-\csc^2x=-\frac1{\sin^2x}\;$$ , we get

$$\int\frac{dx}{\sin^2x\cos^2x}=\frac42\int\frac{d(2x)}{\sin^22x}=-2\cot 2x+C$$

$$\tan(x) - \cot(x) = \tan(x) -\frac{1}{\tan(x)} = \frac{\tan^2(x) - 1}{\tan(x)}$$

As $$\tan(2x) = \frac{2\tan(x)}{1- \tan^2(x)}$$

This implies $$\tan(x) - \cot(x) = -2\cot(2x)$$

Ur correct, actually both r same

\begin{align*} I & = \int\frac{1}{\sin^2(x)\cos^2(x)} \, dx\\ &= \int\frac{\sin^2x+\cos^2x}{\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*}

But this is same as your answer, so nothing wrong with what you arrived at.