Question

Solve:

$\int \frac{1}{{\cos }^{2}x-{\sin }^{2}x}dx$

• A. $\frac{\ln |\sec 2x+\tan 2x|}{2}+C$
• B. $\frac{\ln |\tan 2x|}{2}+C$
• C. $\frac{\ln |\sec 2x|}{2}+C$
• D. None of these

Answer 1

The correct option is A

$\frac{\ln |\sec 2x+\tan 2x|}{2}+C$

Consider the given integral.

$I=\int \frac{1}{{\cos }^{2}x-{\sin }^{2}x}dx$

$I=\int \frac{1}{\cos 2x}dx$

$I=\int \sec 2xdx$

$I=\frac{\ln |\sec 2x+\tan 2x|}{2}+C$

Hence, this is the answer.