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)

$\tan x-\cot x+C$

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