Single Choice

$$\displaystyle \int \frac {dx}{\sin^2 x \cos^2x}$$ equals

A$$\tan x+\cot x+C$$
B$$\tan x-\cot x+C$$
Correct Answer
C$$\tan x \cot x+C$$
D$$\tan x-\cot 2x+C$$

Solution

Let $$\displaystyle I=\int \frac {dx}{\sin^2x\cos^2x}$$
$$\displaystyle =\int \frac {1}{\sin^2x \cos^2x}dx$$
$$\displaystyle =\int \frac {\sin^2x+\cos^2x}{\sin^2x\cos^2x}dx$$
$$\displaystyle =\int \frac {\sin^2x}{\sin^2x\cos^2x}dx+\int \frac {\cos^2x}{\sin^2 x \cos^2x}dx$$
$$\displaystyle =\int \sec^2 x dx+\int co\sec^2 x dx$$
$$=\tan x-\cot x+C$$

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