int frac{d x}{cos (x+4) cos (x+2)} is equal t | vedclass

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(C) Let $I = \int \frac{d x}{\cos (x+4) \cos (x+2)}$.Multiply and divide by $\sin 2$:$I = \frac{1}{\sin 2} \int \frac{\sin 2}{\cos (x+4) \cos (x+2)} d

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{d x}{\cos (x+4) \cos (x+2)}$.Multiply and divide by $\sin 2$:$I = \frac{1}{\sin 2} \int \frac{\sin 2}{\cos (x+4) \cos (x+2)} d x$.Since $2 = (x+4) - (x+2)$,we have:$I = \frac{1}{\sin 2} \int \frac{\sin [(x+4)-(x+2)]}{\cos (x+4) \cos (x+2)} d x$.Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:$I = \frac{1}{\sin 2} \int \frac{\sin (x+4) \cos (x+2) - \cos (x+4) \sin (x+2)}{\cos (x+4) \cos (x+2)} d x$.$I = \frac{1}{\sin 2} \int [	an (x+4) - 	an (x+2)] d x$.Integrating $	an x$,we get $\log |\sec x|$:$I = \frac{1}{\sin 2} [\log |\sec (x+4)| - \log |\sec (x+2)|] + C$.$I = \frac{1}{\sin 2} \log \left|\frac{\sec (x+4)}{\sec (x+2)}ight| + C$.

$\int \frac{d x}{\cos (x+4) \cos (x+2)}$ is equal to

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