Question

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

• A. tan x + cot x +C
• B. tan x - cot x +C
• C. tan x cot x +C
• D. tan x-cot 2x +C

Answer 1

Answer: (B)

tan x - cot x +C

Answer 2

Let$I=\int \frac{dx}{\sin^2 x \cos^2 x}$

= $\int \frac{1}{\sin^2 x \cos^2 x}dx$

= $\int \frac{\sin^2 x+ \cos^2 x}{\sin^2 x \cos^2 x}dx$

= $\int \frac{\sin^2x}{\sin^2x\cos^2 x}dx+\int\frac{\cos^2 x}{\sin^2x\cos^2x}dx$

= $\int \sec^2 xdx+\int\cosec^2 xdx$

= $\tan x -\cot x+C$

Hence, the correct answer is B.