Question

$∫\frac {sin^2 x-cos^2 x}{sin^2 x cos^2 x }\ dx \ is\ equal \ to$

• A. $tan\ x + cot\ x +C$
• B. $tan \ x + cosec \ x +C$
• C. $-tan\ x + cot\ x +C$
• D. $tan \ x + sec\ x +C$

Answer 1

Answer: (A)

$tan\ x + cot\ x +C$

Answer 2

$∫\frac {sin^2 x-cos^2 x}{sin^2 x cos^2 x }\ dx$

= $∫(\frac {sin^2 x}{sin^2 x cos^2 x }\ dx$ - $∫\frac {cos^2 x}{sin^2 x cos^2 x })\ dx$

= $∫(sec^2 x - cosec^2 x) dx$

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

Hence, the correct Answer is (A): $tan\ x + cot \ x +C$