Question:
$\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$ equals
A. $\tan x+\cot x+\mathrm{C}$
B. $\tan x-\cot x+\mathrm{C}$
C. $\tan x \cot x+\mathrm{C}$
D. $\tan x-\cot 2 x+\mathrm{C}$
Solution:
Let $\begin{aligned} I &=\int \frac{d x}{\sin ^{2} x \cos ^{2} x} \\ &=\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x \\ &=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x \\ &=\int \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x \\ &=\int \sec ^{2} x d x+\int \operatorname{cosec}^{2} x d x \\ &=\tan x-\cot x+\mathrm{C} \end{aligned}$
Hence, the correct answer is B.
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