• # question_answer The integral $\int_{{{\tan }^{-1}}\lambda }^{{{\cot }^{-1}}\lambda }{\frac{\tan x}{\tan x+\cot x}dx,}$$\forall \lambda \in R$ cannot take the value A)  $\frac{-\pi }{4}$                               B) $\frac{\pi }{4}$ C)  $\frac{+\pi }{2}$                             D)  $\frac{-3\pi }{4}$

Let $I=\,\int_{{{\tan }^{-1}}}^{{{\cot }^{-1}}}{\frac{\tan \,x}{\tan \,x+\cot \,x}dx}$                      ?(i) $I=\int_{{{\tan }^{-1}}\lambda }^{{{\cot }^{-1}}\lambda }{\frac{\cot \,x}{\cot \,x+\tan \,x}dx}$    ?(ii) $\left( \because \,\,{{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2} \right)$ On adding Eqs. (i) and (ii), we get $2I=\int_{{{\tan }^{-1}}\lambda }^{{{\cot }^{-1}}\lambda }{1\,dx}\,\,\,\,=[x]_{{{\tan }^{-1}}\lambda }^{{{\cot }^{-1}}\lambda }$ $\Rightarrow$            $2I={{\cot }^{-1}}\,\lambda -{{\tan }^{-1}}\lambda$ $\Rightarrow$            $2I=\frac{\pi }{2}\,-{{\tan }^{-1}}\lambda -{{\tan }^{-1}}\lambda$ $\Rightarrow$            $I=\frac{\pi }{4}-{{\tan }^{-1}}\lambda$ $\Rightarrow$            $\frac{-\pi }{2}<{{\tan }^{-1}}\lambda <\frac{\pi }{2}$ $\frac{3\pi }{4}>\frac{\pi }{4}-{{\tan }^{-1}}\,\lambda >-\frac{\pi }{4}$ $\Rightarrow$            $\frac{-\pi }{4}<I<\frac{3\pi }{4}$