Evaluate the integral: I = integral dx/(sin²x cos²x)

Evaluate the integral: $$ I = \int \frac{dx}{\sin^2x\,\cos^2x}. $$

Step 1 · Rewrite the denominator

Since $\sin x\cos x = \tfrac12\sin2x$, squaring gives $\sin^2x\cos^2x = \tfrac14\sin^2(2x)$. Thus:

$$ \frac{1}{\sin^2x\cos^2x} = \frac{4}{\sin^2(2x)} = 4\csc^2(2x). $$

So the integral becomes:

$$ I = 4\int \csc^2(2x)\,dx. $$

Step 2 · Substitution

Let $u=2x$, so $du=2\,dx$:

$$ I = 4\cdot\frac12\int \csc^2u\,du = 2\int \csc^2u\,du. $$

Step 3 · Integrate

Using the standard result $\int \csc^2u\,du = -\cot u$:

$$ I = -2\cot u + C. $$

Step 4 · Substitute back

Substituting back $u=2x$:

$$ I = -2\cot2x + C. $$

Summary Table

Additional Notes

• The domain excludes $x=k\pi/2$ for integer $k$, where either $\sin x$ or $\cos x$ vanishes.
• An equivalent route is to write $\dfrac{1}{\sin^2x\cos^2x}=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\sec^2x+\csc^2x$ and integrate term by term to get $\tan x-\cot x+C$; this differs from $-2\cot2x+C$ only by a constant, since $\tan x-\cot x=-2\cot2x$ identically.
• The result can be checked by differentiating $-2\cot2x$ and confirming it returns $\dfrac{1}{\sin^2x\cos^2x}$.