Question

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

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

$\int \frac{dx}{\cos (x+4)\cos (x+2)}$
Multiplying $\sin 2=\sin \left[\right(x+4)-(x+2\left)\right]$ in both numerator and denominator we get
$\Rightarrow$ $\frac{1}{\sin 2}\int \frac{\sin \left[\right(x+4)-(x+2\left)\right]dx}{\cos (x+4)cos(x+2)}$
$\Rightarrow$ $\frac{1}{\sin 2}\int \frac{\sin (x+4)\cos (x+2)-\sin (x+2)\cos (x+4)dx}{\cos (x+4)\cos (x+2)}$
$\Rightarrow$ $\frac{1}{\sin 2}\int \tan (x+4)-\tan (x+2)$
$\Rightarrow$ $\frac{1}{\sin 2}\left[\log \sec \right(x+4)-\log \sec (x+2\left)\right]+c$ $(as\int \tan \theta =\log secx+c)$
$\Rightarrow$ $\frac{1}{\sin 2}\log |\frac{\sec (x+4)}{\sec (x+2)}|+c$
Therefore Answer is $C$