Question Number 27294 by iy last updated on 04/Jan/18
$$\underset{\mathrm{1}/{e}} {\overset{\mathrm{tan}\:{x}} {\int}}\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:+\:\underset{\mathrm{1}/{e}} {\overset{\mathrm{cot}\:{x}} {\int}}\:\frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:{dt}\:= \\ $$
Commented by prakash jain last updated on 05/Jan/18
![(A) ∫_(1/e) ^(tan x) (t/(1+t^2 ))dt=(1/2)[ln (1+tan^2 x)−ln (1+(1/e^2 ))] =(1/2)[ln (sec^2 x)−ln (1+(1/e^2 ))] =ln (secx)−(1/2)ln (1+(1/e^2 )) ∫_(1/e) ^(cot x) (1/(t(1+t^2 )))dt ∫_(1/e) ^(cot x) ((t/t^2 )−(t/(1+t^2 )))dt =[ln t−(1/2)ln (1+t^2 )]_(1/e) ^(cot x) (B) =ln cot x−(1/2)ln (1+cot^2 x)−ln (1/e)+(1/2)ln (1+(1/e^2 )) =ln cot x−ln cosec x−ln (1/e)+(1/2)ln (1+(1/e^2 )) A+B ln sec x+ln cot x−ln cosec x+1 =ln (sec x∙cot x)−ln cosec x+1 =ln cosec x−ln cosec x+1 =1](https://www.tinkutara.com/question/Q27372.png)
$$\left({A}\right) \\ $$$$\int_{\mathrm{1}/{e}} ^{\mathrm{tan}\:{x}} \frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right)−\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{\mathrm{2}} }\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\left(\mathrm{sec}^{\mathrm{2}} {x}\right)−\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{\mathrm{2}} }\right)\right] \\ $$$$=\mathrm{ln}\:\left(\mathrm{sec}{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{\mathrm{2}} }\right) \\ $$$$\int_{\mathrm{1}/{e}} ^{\mathrm{cot}\:{x}} \frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\int_{\mathrm{1}/{e}} ^{\mathrm{cot}\:{x}} \left(\frac{{t}}{{t}^{\mathrm{2}} }−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt} \\ $$$$=\left[\mathrm{ln}\:{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{1}/{e}} ^{\mathrm{cot}\:{x}} \\ $$$$\left({B}\right) \\ $$$$=\mathrm{ln}\:\mathrm{cot}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} {x}\right)−\mathrm{ln}\:\frac{\mathrm{1}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{ln}\:\mathrm{cot}\:{x}−\mathrm{ln}\:\mathrm{cosec}\:{x}−\mathrm{ln}\:\frac{\mathrm{1}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{\mathrm{2}} }\right) \\ $$$${A}+{B} \\ $$$$\mathrm{ln}\:\mathrm{sec}\:{x}+\mathrm{ln}\:\mathrm{cot}\:{x}−\mathrm{ln}\:\mathrm{cosec}\:{x}+\mathrm{1} \\ $$$$=\mathrm{ln}\:\left(\mathrm{sec}\:{x}\centerdot\mathrm{cot}\:{x}\right)−\mathrm{ln}\:\mathrm{cosec}\:{x}+\mathrm{1} \\ $$$$=\mathrm{ln}\:\mathrm{cosec}\:{x}−\mathrm{ln}\:\mathrm{cosec}\:{x}+\mathrm{1} \\ $$$$=\mathrm{1} \\ $$