MCQ
Choose the correct answer in Exercises:
$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$ is equal to
  • $\tan\text{x}+\cot\text{x}+\text{C}$
  • B
    $\tan\text{x}+\text{cosec x}+\text{C}$
  • C
    $-\tan\text{x}+\cot\text{x}+\text{C}$
  • D
    $\tan\text{x}+\sec\text{x}+\text{C}$

Answer

Correct option: A.
$\tan\text{x}+\cot\text{x}+\text{C}$
$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\bigg(\frac{\sin^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}-\frac{\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int(\sec^2\text{x}-\text{cosec}^2\text{x})\text{ dx}$
$=\tan\text{x}+\cot\text{x}+\text{C}$
Hence, the correct answer is A.

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