int sec left(x-frac{pi}{3}right) sec left(x+f | vedclass

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Question

(B) Let $I = \int \sec \left(x-\frac{\pi}{3}\right) \sec \left(x+\frac{\pi}{6}\right) d x$. Multiply and divide by $\sin \left(\left(x+\frac{\pi}{6}\r

Vedclass · Accepted Answer

$\log \left|\frac{\cos \left(x-\frac{\pi}{3}\right)}{\cos \left(x+\frac{\pi}{6}\right)}\right|+c$

Vedclass · Answer

(B) Let $I = \int \sec \left(x-\frac{\pi}{3}ight) \sec \left(x+\frac{\pi}{6}ight) d x$. Multiply and divide by $\sin \left(\left(x+\frac{\pi}{6}ight) - \left(x-\frac{\pi}{3}ight)ight) = \sin \left(\frac{\pi}{6} + \frac{\pi}{3}ight) = \sin \left(\frac{\pi}{2}ight) = 1$. So,$I = \int \frac{\sin \left(\left(x+\frac{\pi}{6}ight) - \left(x-\frac{\pi}{3}ight)ight)}{\cos \left(x-\frac{\pi}{3}ight) \cos \left(x+\frac{\pi}{6}ight)} d x$. Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we get: $I = \int \frac{\sin \left(x+\frac{\pi}{6}ight) \cos \left(x-\frac{\pi}{3}ight) - \cos \left(x+\frac{\pi}{6}ight) \sin \left(x-\frac{\pi}{3}ight)}{\cos \left(x-\frac{\pi}{3}ight) \cos \left(x+\frac{\pi}{6}ight)} d x$. $I = \int \left( \frac{\sin \left(x+\frac{\pi}{6}ight)}{\cos \left(x+\frac{\pi}{6}ight)} - \frac{\sin \left(x-\frac{\pi}{3}ight)}{\cos \left(x-\frac{\pi}{3}ight)} ight) d x$. $I = \int 	an \left(x+\frac{\pi}{6}ight) d x - \int 	an \left(x-\frac{\pi}{3}ight) d x$. $I = \log \left| \sec \left(x+\frac{\pi}{6}ight) ight| - \log \left| \sec \left(x-\frac{\pi}{3}ight) ight| + c$. $I = \log \left| \frac{\sec \left(x+\frac{\pi}{6}ight)}{\sec \left(x-\frac{\pi}{3}ight)} ight| + c$. Note: The options provided in the question are slightly different in form. However,using $\sec 	heta = 1/\cos 	heta$,we can write this as $\log \left| \frac{\cos \left(x-\frac{\pi}{3}ight)}{\cos \left(x+\frac{\pi}{6}ight)} ight| + c$,which matches option $B$.

$\int \sec \left(x-\frac{\pi}{3}\right) \sec \left(x+\frac{\pi}{6}\right) d x=$

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