Question

Evaluate: $\int \frac{dx}{\cos (x+4)\cos (x+2)}$

• A. $\frac{1}{\sin 2}\log \left|\cos {(x+4)}^2\right|+c$
• B. $\frac{1}{2}\log \left|\frac{\sec (x+2)}{\sec (x+4)}\right|+c$
• C. $\frac{1}{\sin 2}\log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+c$
• D. $\log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+c$

Answer 1

The correct option is C $\frac{1}{\sin 2}\log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+c$

Consider, $I=\int \frac{dx}{\cos (x+4).\cos (x+2)}$

$I=\frac{1}{\sin 2}\int \frac{dx.\sin 2}{\cos (x+4).\cos (x+2)}$

$I=\frac{1}{\sin 2}.\int \frac{dx.\sin [x-x+4-2]}{\cos (x+4).\cos (x+2)}$

$I=\frac{1}{\sin 2}.\int \frac{dx.\sin [(x+4)-(x+2)]}{\cos (x+4).\cos (x+2)}$

$I=\frac{1}{\sin 2}\times [\int \tan (x+4)dx-\int \tan (x+2)dx]$

$I=\frac{1}{\sin 2}\times (\log \left|\sec (x+4)\right|-\log \left|\sec (x+2)\right|)$

$I=\frac{1}{\sin 2}\times \log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+c$