RIGHT)dx = m\left( {\pi + n} \right)\) The given integral is:\(I = \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \left( {\frac{{cotx}}{{cotx + cscx}}} \right)dx\) Now,\(I = \mathop \smallint \limits_0^{\pi /2} \left( {\frac{{\frac{{\cos x}}{{\sin x}}}}{{\frac{{\cos x}}{{\sin {\rm{x}}}} + \frac{1}{{\sin x}}}}} \right)dx\) \(\because \left[ \begin{matrix}\cot x=\frac{\cos x}{\sin x} \\cosecx=\frac{1}{\sin x} \\\end{matrix} \right]\) \(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\pi /2} \left( {\frac{{\frac{{\cos x}}{{\sin x}}}}{{\frac{{\cos x + 1}}{{\sin x}}}}} \right)dx\) \(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\pi /2} \left( {\frac{{\cos x}}{{\sin x}} \times \frac{{\sin x}}{{\cos x + 1}}} \right)dx\) \(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\pi /2} \left( {\frac{{\cos x}}{{\cos x + 1}}} \right)dx\) \(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\pi /2} \left( {\frac{{2{{\cos }^2}\left( {\frac{x}{2}} \right) - 1}}{{2{{\cos }^2}\left( {\frac{x}{2}} \right)}}} \right)dx\) \(\because \left[ \begin{matrix}\cos \left( x \right)=2{{\cos }^{2}}\left( \frac{x}{2} \right)-1 \\\Rightarrow \cos \left( x \right)+1=2{{\cos }^{2}}\left( \frac{x}{2} \right) \\\end{matrix} \right]\) \(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\pi /2} \left( {\frac{{2{{\cos }^2}\left( {\frac{x}{2}} \right)\left[ {1 - \frac{1}{{2{{\cos }^2}\left( {\frac{x}{2}} \right)}}} \right]}}{{2{{\cos }^2}\left( {\frac{x}{2}} \right)}}} \right)d{\rm{x}}\) \(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\pi /2} \left( {1 - \frac{1}{{2{{\cos }^2}\left( {\frac{x}{2}} \right)}}} \right)dx\) \(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\pi /2} \left( {1 - \left( {\frac{1}{2}{{\sec }^2}\left( {\frac{x}{2}} \right)} \right)} \right)dx\) On integrating,\(\Rightarrow {\rm{I}} = {\rm{}}\left[ x \right]_0^{\pi /2}{\rm{}} - \mathop \smallint \limits_0^{\pi /2} \left( {\frac{1}{2}{{\sec }^2}\left( {\frac{x}{2}} \right)} \right)dx\) PUTTING \(u = \frac{x}{2}\)\(\Rightarrow \frac{{du}}{{dx}} = \frac{1}{2}\) ⇒ 2du = dx\(\Rightarrow {\rm{I}} = {\rm{\;}}\left[ x \right]_0^{\pi /2} - \mathop \smallint \limits_0^{\pi /2} \left( {\frac{1}{2}{{\sec }^2}\left( u \right)} \right)2du\) \(\Rightarrow {\rm{I}} = {\rm{\;}}\left[ x \right]_0^{\pi /2} - \mathop \smallint \limits_0^{\pi /2} \left( {{{\sec }^2}\left( u \right)} \right)du\) ∵ \(\left[ {\smallint {{\sec }^2}\left( u \right)du = \tan \left( u \right)} \right]\)\(\Rightarrow {\rm{I}} = {\rm{\;}}\left[ x \right]_0^{\pi /2} - \left[ {\tan u} \right]_0^{\pi /2}\) \(\Rightarrow {\rm{I}} = {\rm{\;}}\left[ x \right]_0^{\pi /2} - \left[ {\tan \left( {\frac{x}{2}} \right)} \right]_0^{\pi /2}\) On APPLYING limits,\(\Rightarrow {\rm{I}} = {\rm{\;}}\frac{\pi }{2} - 0 - \tan \frac{\pi }{4} + \tan 0\) ∵ \(\left[ {\tan \frac{\pi }{4} = 1{\rm{\;and\;}}\tan 0 = 0} \right]\)\(\Rightarrow {\rm{I}} = \frac{{\rm{\pi }}}{2} - 1 + 0\) \(\Rightarrow {\rm{I}} = {\rm{\;}}\frac{\pi }{2} - 1\) Thus,\(\Rightarrow \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \left( {\frac{{cotx}}{{cotx + cosecx}}} \right)dx = \frac{\pi }{2} - 1 = m\left( {\pi + n} \right)\) \(\Rightarrow \frac{1}{2}\left( {\pi - 2} \right) = m\left( {\pi + n} \right)\) \(\Rightarrow \frac{1}{2}\left( {\pi + \left( { - 2} \right)} \right) = m\left( {\pi + n} \right)\) On COMPARING,\(\Rightarrow m = \frac{1}{2}\) ⇒ n = -2Now,\(\Rightarrow m \cdot n = \frac{1}{2} \times \left( { - 2} \right)\) ∴ m⋅n = -1