Integral Calculus
86338
\(\int \frac{1}{\cos (x+4) \cos (x+2)} d x\) is equal to
1 \(\frac{1}{\sin 2} \log \left|\cos (\mathrm{x}+4)^{2}\right|+\mathrm{c}\)
2 \(\frac{1}{2} \log \left|\frac{\sec (\mathrm{x}+2)}{\sec (\mathrm{x}+4)}\right|+\mathrm{c}\)
3 \(\frac{1}{\sin 2} \log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+c\)
4 \(\log \left|\frac{\sec (\mathrm{x}+4)}{\sec (\mathrm{x}+2)}\right|+\mathrm{c}\)
Explanation:
(C) : Given,
\(I=\int \frac{1}{\cos (x+4) \cos (x+2)} d x\)
\(I=\frac{1}{\sin 2} \int \frac{\sin 2}{\cos (x+4) \cdot \cos (x+2)} d x\)
\(I=\frac{1}{\sin 2} \int \frac{\sin [x-x+4-2]}{\cos (x+4) \cdot \cos (x+2)} d x\)
\(I=\frac{1}{\sin 2} \int \frac{\sin [(x+4)-(x+2)]}{\cos (x+4) \cdot \cos (x+2)} d x\)
\(I=\frac{1}{\sin 2}\left[\int \tan (x+4) d x-\int \tan (x+2) d x\right]\)
\(I=\frac{1}{\sin 2}[\log |\sec (x+4)|-\log |\sec (x+2)|]+c\)
\(I=\frac{1}{\sin 2} \times \log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+c\)