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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} \int \frac{\mathrm{d} x}{\cot ^2 x-1} & =\int \frac{\mathrm{d} x}{\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x}} \ & =\int \frac{\sin ^2 x}{\cos 2 x} \mathrm{~d} x \ & =\int \frac{\frac{1-\cos 2 x}{2}}{\cos 2 x} \mathrm{~d} x \ & =\frac{1}{2} \int(\sec 2 x-1) \mathrm{d} x \ & =\frac{1}{2}\left(\frac{\log |\sec 2 x+	an 2 x|}{2}-xight)+\mathrm{c} \ & =\frac{1}{4} \log |\sec 2 x+	an 2 x|-\frac{x}{2}+\mathrm{c} \ 	herefore \quad \mathrm{A}=4, \mathrm{~B} & =2 \ \Rightarrow \mathrm{A}+\mathrm{B} & =6\end{aligned}$

$\int \frac{\mathrm{d} x}{\cot ^2 x-1}=\frac{1}{\mathrm{~A}} \log |\sec 2 x+\tan 2…

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