Question Number 101378 by Rohit@Thakur last updated on 02/Jul/20

∫_(1/e) ^(tanx) (t/(1+t^2 ))dt + ∫_(1/e) ^(cotx) (1/(t(1+t^2 )))dt

Commented byDwaipayan Shikari last updated on 02/Jul/20

(1/2)∫_(1/e) ^(tanx) ((2tdt)/(1+t^2 ))=[(1/2)log(1+t^2 )]_(1/e) ^(tanx) =(1/2)log(sec^2 x)−(1/2)log(((1+e^2 )/e^2 )) −(1/2)∫_(1/e) ^(cotx) (((−2dt)/t^3 )/(1+(1/t^2 )))=[−(1/2)log(1+(1/t^2 ))]_(1/e) ^(cotx) =−(1/2)log(((cosec^2 x)/(cot^2 x)))+(1/2)log(1+e^2 ) =−(1/2)log(sec^2 x)+(1/2)log(1+e^2 ) So ∫_(1/e) ^(tanx) (t/(1+t^2 ))+∫_(1/e) ^(cotx) (1/(t(1+t^2 )))=(1/2)log(1+e^2 )−(1/2)log(((1+e^2 )/e^2 )) =(1/2)loge^2 =1 ★■L

Answered by 1549442205 last updated on 02/Jul/20

A=∫(t/(1+t^2 ))dt=(1/2)∫((d(1+t^2 ))/(1+t^2 ))=(1/2)ln(1+t^2 ) B=∫(dt/(t(1+t^2 ))).Putting t=tan u⇒dt ⇒ { ((dt=(1+t^2 )du)),((sin u=(√(1/(1+cot^2 u)))=(1/(√(1+(1/t^2 ))))=(t/(√(1+t^2 ))))) :} ⇒B=∫(du/(tan u))=∫((cos u)/(sin u))du=∫((dsin u)/(sin u)) =ln∣sin u∣=ln∣(t/(√(1+t^2 )))∣Hence, ∫_(1/e) ^(tanx) (t/(1+t^2 ))dt + ∫_(1/e) ^(cotx) (1/(t(1+t^2 )))dt= A(tan x)−A((1/e))+B(cotx)−B((1/e)) =(1/2)ln(1/(cos^2 x))−(1/2)ln(1+e^(−2) )+ln∣cos x∣ −ln(1/(√(1+e^2 )))=−(1/2)ln(1+e^(−2) )+ln(√(1+e^2 )) =ln((√(1+e^2 ))/(√(1+e^(−2) )))=lne=1

Answered by mathmax by abdo last updated on 02/Jul/20

we have ∫_(1/e) ^(tanx ) (t/(1+t^2 ))dt =[(1/2)ln(1+t^2 )]_(1/e) ^(tanx) =(1/2)ln(1+tan^2 x)−(1/2)ln(1+e^(−2) ) =(1/2)ln((1/(cos^2 x)))−(1/2)ln(1+e^(−2) ) =−ln∣cosx∣ −(1/2)ln(1+e^(−2) ) ∫_(1/e) ^(cotanx ) (dt/(t(1+t^2 ))) =∫_(1/e) ^(1/(tanx)) ((1/t)−(t/(1+t^2 )))dt =[ln∣t∣]_(1/e) ^(1/(tanx)) −[(1/2)ln(1+t^2 )]_(1/e) ^(1/(tanx)) =−ln∣tanx∣+1 −(1/2)ln(1+(1/(tan^2 x)))−(1/2)ln(1+e^(−2) ) =−ln∣tanx∣−(1/2)ln(1+tan^2 x)+ln∣tanx∣ −(1/2)ln(1+e^(−2) )+1 ⇒ ∫_(1/e) ^(tanx ) ((tdt)/(1+t^2 )) +∫_(1/e) ^(cotanx) (dt/(t(1+t^2 ))) = (1/2)ln(1+tan^2 x)−(1/2)ln(1+e^(−2) ) −(1/2)ln(1+tan^2 x)−(1/2)ln(1+e^(−2) ) +1 =1