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$\frac{\pi}{2}$
If $\int_0 \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=m(\pi+n)$, then (m.n) equals
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If $\int_0 \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=m(\pi+n)$, then (m.n) equals
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The correct answer is:
$-1$
$\begin{aligned} & \int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=\int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x+1} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{2 \cos ^2 \frac{x}{2}-1}{2 \cos ^2 \frac{x}{2}} d x=\int_0^{\frac{\pi}{2}}\left(1-\frac{1}{2} \sec ^2 \frac{x}{2}\right) d x \\ & =\left[x-\tan \frac{x}{2}\right]_0^{\pi / 2}=\frac{\pi}{2}-1=\frac{1}{2}(\pi+(-2)) \\ & \Rightarrow m=\frac{1}{2} \text { and } n=-2 \\ & \Rightarrow \mathrm{m} \times \mathrm{n}=\frac{1}{2} \times(-2)=-1\end{aligned}$
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