MCQ
Choose the correct answer in Exercises:
$\int\frac{\text{dx}}{\sin^2\text{x}\cos^2\text{x}}\text{ equals}$
  • A
    $\tan\text{x}+\cot\text{x}+\text{C}$
  • $\tan\text{x}-\cot\text{x}+\text{C}$
  • C
    $\tan\text{x}\cot\text{x}+\text{C}$
  • D
    $\tan\text{x}-\cot\text{2x}+\text{C}$

Answer

Correct option: B.
$\tan\text{x}-\cot\text{x}+\text{C}$
$\int\frac{\text{dx}}{\sin^2\text{x}\cos^2\text{x}}= \int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{\sin^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}+ \frac{\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{1}{\cos^2\text{x}}+\frac{1}{\sin^2\text{x}}\text{ dx}=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x}\text{ dx}$
$=\int\sec^2\text{x}\text{ dx} +\text{ dx}\int\ \text{cosec}^2\text{x}\text{ dx}$
$=\tan\text{x}-\cot\text{x}+\text{c} $
Therefore, option (B) is correct.

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