Question:
$\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$ is equal to
A. $\tan x+\cot x+C$
B. $\tan x+\operatorname{cosec} x+C$
C. $-\tan x+\cot x+C$
D. $\tan x+\sec x+C$
Solution:
$\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x=\int\left(\frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\right) d x$
$=\int\left(\sec ^{2} x-\operatorname{cosec}^{2} x\right) d x$
$=\tan x+\cot x+\mathrm{C}$
Hence, the correct answer is A.