If int frac{d x}{cot ^2 x-1}=frac{1}{A} log | | vedclass

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(B) We are given the integral $I = \int \frac{d x}{\cot ^2 x-1}$.First,rewrite the integrand using $\cot x = \frac{\cos x}{\sin x}$:$I = \int \frac{d

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$6$

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(B) We are given the integral $I = \int \frac{d x}{\cot ^2 x-1}$.First,rewrite the integrand using $\cot x = \frac{\cos x}{\sin x}$:$I = \int \frac{d x}{\frac{\cos ^2 x}{\sin ^2 x}-1} = \int \frac{\sin ^2 x}{\cos ^2 x-\sin ^2 x} d x$.Using the identity $\cos 2x = \cos ^2 x - \sin ^2 x$,we get:$I = \int \frac{\sin ^2 x}{\cos 2 x} d x$.Using the identity $\sin ^2 x = \frac{1-\cos 2 x}{2}$,we have:$I = \int \frac{1-\cos 2 x}{2 \cos 2 x} d x = \frac{1}{2} \int (\sec 2 x - 1) d x$.Integrating term by term:$I = \frac{1}{2} \left( \frac{\log |\sec 2 x + 	an 2 x|}{2} - x ight) + c$.$I = \frac{1}{4} \log |\sec 2 x + 	an 2 x| - \frac{x}{2} + c$.Comparing this with the given form $\frac{1}{A} \log |\sec 2 x + 	an 2 x| - \frac{x}{B} + c$,we find $A = 4$ and $B = 2$.Therefore,$A + B = 4 + 2 = 6$.

If $\int \frac{d x}{\cot ^2 x-1}=\frac{1}{A} \log |\sec 2 x+\tan 2 x|-\frac{x}{B}+c$,(where $c$ is the constant of integration),then $A+B=$

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