If int frac{1}{cot frac{x}{2} cot frac{x}{3} | vedclass

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(A) We are given the integral $I = \int 	an \frac{x}{2} 	an \frac{x}{3} 	an \frac{x}{6} dx$. Using the identity $\frac{x}{2} = \frac{x}{3} + \frac{

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$7$

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(A) We are given the integral $I = \int 	an \frac{x}{2} 	an \frac{x}{3} 	an \frac{x}{6} dx$. Using the identity $\frac{x}{2} = \frac{x}{3} + \frac{x}{6}$,we have $	an \frac{x}{2} = 	an \left( \frac{x}{3} + \frac{x}{6} ight) = \frac{	an \frac{x}{3} + 	an \frac{x}{6}}{1 - 	an \frac{x}{3} 	an \frac{x}{6}}$. This implies $	an \frac{x}{2} (1 - 	an \frac{x}{3} 	an \frac{x}{6}) = 	an \frac{x}{3} + 	an \frac{x}{6}$. Rearranging,we get $	an \frac{x}{2} - 	an \frac{x}{2} 	an \frac{x}{3} 	an \frac{x}{6} = 	an \frac{x}{3} + 	an \frac{x}{6}$. Therefore,$	an \frac{x}{2} 	an \frac{x}{3} 	an \frac{x}{6} = 	an \frac{x}{2} - 	an \frac{x}{3} - 	an \frac{x}{6}$. Now,integrate each term: $I = \int (	an \frac{x}{2} - 	an \frac{x}{3} - 	an \frac{x}{6}) dx$. The integral of $	an(ax)$ is $-\frac{1}{a} \log |\cos(ax)|$. $I = -2 \log |\cos \frac{x}{2}| + 3 \log |\cos \frac{x}{3}| + 6 \log |\cos \frac{x}{6}| + k$. Comparing with $A \log |\cos \frac{x}{2}| + B \log |\cos \frac{x}{3}| + C \log |\cos \frac{x}{6}| + k$,we get $A = -2$,$B = 3$,$C = 6$. Thus,$A + B + C = -2 + 3 + 6 = 7$.

If $\int \frac{1}{\cot \frac{x}{2} \cot \frac{x}{3} \cot \frac{x}{6}} d x=A \log \left|\cos \frac{x}{2}\right|+B \log \left|\cos \frac{x}{3}\right|+C \log \left|\cos \frac{x}{6}\right|+k$ then $A+B+C=$

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