Question

$\int \frac{1}{\sin x\cos x}$ dx is equal to

• A. $\log |\tan x|+C$
• B. $\log |\sin 2x|+C$
• C. $\log |\sec x|+C$
• D. $\log |\cos x|+C$
• E. $\log |\sin x|+C$

Answer 1

Answer: (A)

Let $I=\int \left(\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right)dx$
$=\int \left(\frac{{\sin }^{2}x}{\sin x\cos x}+\frac{{\cos }^{2}x}{\sin x\cos x}dx\right)$
$=\int (\tan x+\cot x)dx$
$=\log \sec x+(-\log \mathrm{cosec}x)+C$
$=\log \left|\frac{\sec x}{\mathrm{cosec}x}\right|+C=\log |\tan x|+C$