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Question Number 40153 by maxmathsup by imad last updated on 16/Jul/18
calculate ∫_1 ^2 ((t−2)/( (√(t^2 −1))))dt
calculate∫12t−2t2−1dt
Commented by maxmathsup by imad last updated on 17/Jul/18
I = ∫_1 ^2 ((tdt)/( (√(t^2 −1))))dt −2 ∫_1 ^2 (dt/( (√(t^2 −1)))) but ∫_1 ^2 (t/( (√(t^2 −1))))dt =[(√(t^2 −1))]_1 ^2 =(√3) also changement t=chu give ∫ (dt/( (√(t^2 −1)))) =∫ ((shu du)/(shu)) = u +c = argch(t)=ln(t+(√(t^2 −1)))+c ⇒ ∫_1 ^2 (dt/( (√(t^2 −1)))) =[ln(t+(√(t^2 −1)))]_1 ^2 =ln(2+(√3)) ⇒ I =(√3) −2ln(2+(√3)) .
I=∫12tdtt2−1dt−2∫12dtt2−1but∫12tt2−1dt=[t2−1]12=3alsochangementt=chugive∫dtt2−1=∫shudushu=u+c=argch(t)=ln(t+t2−1)+c⇒∫12dtt2−1=[ln(t+t2−1)]12=ln(2+3)⇒I=3−2ln(2+3).
Answered by sma3l2996 last updated on 16/Jul/18
∫_1 ^2 ((t−2)/( (√(t^2 −1))))dt=∫_1 ^2 (t/( (√(t^2 −1))))dt−2∫_1 ^2 (dt/( (√(t^2 −1)))) =[(√(t^2 −1))]_1 ^2 −2∫_1 ^2 ((t+(√(t^2 −1)))/((t+(√(t^2 −1)))(√(t^2 −1))))dt =(√5)−2∫_1 ^2 ((t+(√(t^2 −1)))/( (√(t^2 −1))))×(dt/(t+(√(t^2 −1)))) =(√5)−2∫_1 ^2 ((t/( (√(t^2 −1))))+1)×(1/(t+(√(t^2 −1))))dt =(√5)−2∫_1 ^2 ((d(t+(√(t^2 −1))))/(t+(√(t^2 −1)))) =(√5)−2[ln∣t+(√(t^2 −1))∣]_1 ^2 =(√5)−2ln(2+(√5))
∫12t−2t2−1dt=∫12tt2−1dt−2∫12dtt2−1=[t2−1]12−2∫12t+t2−1(t+t2−1)t2−1dt=5−2∫12t+t2−1t2−1×dtt+t2−1=5−2∫12(tt2−1+1)×1t+t2−1dt=5−2∫12d(t+t2−1)t+t2−1=5−2[ln∣t+t2−1∣]12=5−2ln(2+5)

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