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Question Number 131201 by shaker last updated on 02/Feb/21

Answered by liberty last updated on 11/Feb/21

L=∫ ((tan^2 x)/(cos 2x)) dx ; by change of variable let u = tan x → { ((dx=(du/(1+u^2 )))),((cos 2x=((1−u^2 )/(1+u^2 )))) :} L=∫ (((u^2 (1+u^2 ))/(1−u^2 )))((du/(1+u^2 ))) L=∫ (u^2 /(1−u^2 )) du = ∫ (−1−(1/(1−u^2 )))du L=−u−(1/2)∫ [ (1/(1−u))+(1/(1+u)) ]du L=−u−(1/2)ln ∣((1+u)/(1−u)) ∣+c L=−tan x−(1/2)ln ∣((tan x+1)/(1−tan x)) ∣ + c

L=∫tan2xcos2xdx;bychangeofvariableletu=tanx→{dx=du1+u2cos2x=1−u21+u2L=∫(u2(1+u2)1−u2)(du1+u2)L=∫u21−u2du=∫(−1−11−u2)duL=−u−12∫[11−u+11+u]duL=−u−12ln∣1+u1−u∣+cL=−tanx−12ln∣tanx+11−tanx∣+c

Answered by mathmax by abdo last updated on 02/Feb/21

I =∫ ((tan^2 x)/(cos(2x)))dx ⇒I =∫(((1+tan^2 x)tan^2 x)/(1−tan^2 x))dx ⇒ I =_(tanx=t) ∫ ((t^2 (1+t^2 ))/((1−t^2 )))(dt/(1+t^2 )) =∫ (t^2 /(1−t^2 ))dt =−∫ (t^2 /(t^2 −1))dt =−∫ ((t^2 −1+1)/(t^2 −1))dt =−t−∫ (dt/((t−1)(t+1))) =−t −(1/2)∫ ((1/(t−1))−(1/(t+1)))dt=−t−(1/2)ln∣((t−1)/(t+1))∣ +C =−tanx−(1/2)ln∣((tanx−1)/(tanx+1))∣+C

I=∫tan2xcos(2x)dx⇒I=∫(1+tan2x)tan2x1−tan2xdx⇒I=tanx=t∫t2(1+t2)(1−t2)dt1+t2=∫t21−t2dt=−∫t2t2−1dt=−∫t2−1+1t2−1dt=−t−∫dt(t−1)(t+1)=−t−12∫(1t−1−1t+1)dt=−t−12ln∣t−1t+1∣+C=−tanx−12ln∣tanx−1tanx+1∣+C

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