Menu Close

let-A-0-pi-dx-cosx-sinx-R-1-find-a-explicit-form-of-A-2-find-also-B-0-pi-dx-cosx-sinx-2-3-calculate-0-pi-dx-2-cosx-sinx-and-0-pi-dx

Tinku Tara 0 Comments
Pin




Question Number 64850 by mathmax by abdo last updated on 22/Jul/19
let A_λ =∫_0 ^π (dx/(λ +cosx +sinx)) (λ ∈ R) 1) find a explicit form of A_λ 2) find also B_λ =∫_0 ^π (dx/((λ +cosx +sinx)^2 )) 3) calculate ∫_0 ^π (dx/(2+cosx +sinx)) and ∫_0 ^π (dx/((3+cosx +sinx)^2 ))
letAλ=∫0πdxλ+cosx+sinx(λ∈R)1)findaexplicitformofAλ2)findalsoBλ=∫0πdx(λ+cosx+sinx)23)calculate∫0πdx2+cosx+sinxand∫0πdx(3+cosx+sinx)2
Commented by mathmax by abdo last updated on 23/Jul/19
1) A_λ =∫_0 ^π (dx/(λ +cosx +sinx)) changement tan((x/2))=t give A_λ =∫_0 ^∞ (1/(λ +((1−t^2 )/(1+t^2 )) +((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^∞ ((2dt)/(λ+λt^2 +1−t^2 +2t)) =2∫_0 ^∞ (dt/((λ−1)t^2 +2t +1+λ)) let decompose F(t)=(1/((λ−1)t^2 +2t+1+λ)) Δ^′ =1−(λ−1)(λ+1) =1−(λ^2 −1) =2−λ^2 case 1 2−λ^2 >0 ⇒∣λ∣<(√2) ⇒t_1 =((−1+(√(2−λ^2 )))/(λ−1)) (λ≠1) t_2 =((−1−(√(2−λ^2 )))/(λ−1)) ⇒F(t) =(1/((λ−1)(t−t_1 )(t−t_2 ))) ⇒ A_λ =(2/(λ−1))∫_0 ^∞ (1/((2(√(2−λ^2 )))/(λ−1))) ((1/(t−t_1 )) −(1/(t−t_2 )))dt =(1/( (√(2−λ^2 ))))∫_0 ^∞ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/(2(√(2−λ^2 ))))[ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^(+∞) =(1/(2(√(2−λ^2 ))))ln∣(t_2 /t_1 )∣ A_λ =(1/(2(√(2−λ^2 ))))ln∣((1+(√(2−λ^2 )))/(1−(√(2−λ^2 ))))∣
1)Aλ=∫0πdxλ+cosx+sinxchangementtan(x2)=tgiveAλ=∫0∞1λ+1−t21+t2+2t1+t22dt1+t2=∫0∞2dtλ+λt2+1−t2+2t=2∫0∞dt(λ−1)t2+2t+1+λletdecomposeF(t)=1(λ−1)t2+2t+1+λΔ′=1−(λ−1)(λ+1)=1−(λ2−1)=2−λ2case12−λ2>0⇒∣λ∣<2⇒t1=−1+2−λ2λ−1(λ≠1)t2=−1−2−λ2λ−1⇒F(t)=1(λ−1)(t−t1)(t−t2)⇒Aλ=2λ−1∫0∞122−λ2λ−1(1t−t1−1t−t2)dt=12−λ2∫0∞(1t−t1−1t−t2)dt=122−λ2[ln∣t−t1t−t2∣]0+∞=122−λ2ln∣t2t1∣Aλ=122−λ2ln∣1+2−λ21−2−λ2∣
Commented by mathmax by abdo last updated on 23/Jul/19
case 2 2−λ^2 <0 ⇒∣λ∣>(√2) ⇒Δ^′ <0 and (λ−1)t^2 +2t +1+λ =(λ−1){t^2 +(2/(λ−1))t +((1+λ)/(λ−1))} =(λ−1){ t^2 +((2t)/(λ−1)) +(1/((λ−1)^2 )) +((1+λ)/(λ−1)) −(1/((λ−1)^2 ))} =(λ−1){ (t +(1/(λ−1)))^2 +((λ^2 −2)/((λ−1)^2 ))} we use the changement t+(1/(λ−1)) =((√(λ^2 −2))/(∣λ−1∣)) u ⇒A_λ =(2/(λ−1))∫_0 ^∞ (dt/((t+(1/(λ−1)))^2 +((λ^2 −2)/((λ−1)^2 )))) =(2/(λ−1))∫_((s(λ))/( (√(λ^2 −2)))) ^(+∞) (1/(((λ^2 −2)/((λ−1)^2 ))(1+u^2 )))((√(λ^2 −2))/(∣λ−1∣)) du =((2(λ−1)^2 )/((λ−1)∣λ−1∣)) (√(λ^2 −2)) [arctanu]_((s(λ))/( (√(λ^2 −2)))) ^(+∞) A_λ =2s(λ−1)(√(λ^2 −2)) ((π/2) −arctan(((s(λ))/( (√(λ^2 −2)))))) with s(x) =1 if x>0 and s(x)=−1 if x<0
case22−λ2<0⇒∣λ∣>2⇒Δ′<0and(λ−1)t2+2t+1+λ=(λ−1){t2+2λ−1t+1+λλ−1}=(λ−1){t2+2tλ−1+1(λ−1)2+1+λλ−1−1(λ−1)2}=(λ−1){(t+1λ−1)2+λ2−2(λ−1)2}weusethechangementt+1λ−1=λ2−2∣λ−1∣u⇒Aλ=2λ−1∫0∞dt(t+1λ−1)2+λ2−2(λ−1)2=2λ−1∫s(λ)λ2−2+∞1λ2−2(λ−1)2(1+u2)λ2−2∣λ−1∣du=2(λ−1)2(λ−1)∣λ−1∣λ2−2[arctanu]s(λ)λ2−2+∞Aλ=2s(λ−1)λ2−2(π2−arctan(s(λ)λ2−2))withs(x)=1ifx>0ands(x)=−1ifx<0
Commented by mathmax by abdo last updated on 23/Jul/19
2) we have (dA_λ /dλ) =−∫_0 ^π (dx/((λ +cosx +sinx)^2 )) =−B_λ ⇒ B_λ =−A_λ ^′ rest to calculate A_λ ^′ ....
2)wehavedAλdλ=−∫0πdx(λ+cosx+sinx)2=−Bλ⇒Bλ=−Aλ′resttocalculateAλ′….
Commented by mathmax by abdo last updated on 23/Jul/19
3) let calculate ∫_0 ^π (dx/(2+cosx +sinx)) here λ =2 and 2−λ^2 =−2<0 ⇒∫_0 ^π (dx/(2+cosx +sinx)) =A_2 =2s(1)(√(2^2 −2)){ (π/2) −arctan(((s(2))/( (√(2^2 −2)))))} =2(√2){(π/2) −arctan((1/( (√2))))} =π(√2) −2(√2)((π/2) −arctan((√2))) =π(√2)−π(√2) +2(√2)arctan((√2)) ⇒ ∫_0 ^π (dx/(2+cosx +sinx)) =2(√2)arctan((√2)) .
3)letcalculate∫0πdx2+cosx+sinxhereλ=2and2−λ2=−2<0⇒∫0πdx2+cosx+sinx=A2=2s(1)22−2{π2−arctan(s(2)22−2)}=22{π2−arctan(12)}=π2−22(π2−arctan(2))=π2−π2+22arctan(2)⇒∫0πdx2+cosx+sinx=22arctan(2).

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *