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Question Number 84578 by msup trace by abdo last updated on 14/Mar/20
calculate ∫_0 ^(π/4) (dx/((cosx +3sinx)^2 ))
calculate∫0π4dx(cosx+3sinx)2
Commented by jagoll last updated on 14/Mar/20
∫ _0 ^(π/4) (( dx)/(((√(10)) cos (x−tan^(−1) (3))^2 )) = (1/(10)) ∫ _0 ^(π/4) sec^2 (x−tan^(−1) (3))dx =(1/(10)) tan (x−tan^(−1) (3))] _0^(π/4) = (1/(10)) [tan ((π/4)−tan^(−1) (3))−tan (−tan^(−1) (3))]
∫π40dx(10cos(x−tan−1(3))2=110∫π40sec2(x−tan−1(3))dx=110tan(x−tan−1(3))]0π4=110[tan(π4−tan−1(3))−tan(−tan−1(3))]
Commented by jagoll last updated on 14/Mar/20
=(1/(10)) [((1−3)/(1+3)) +3 ] = (1/(10)) [(5/2)] = (1/4)
=110[1−31+3+3]=110[52]=14
Commented by abdomathmax last updated on 14/Mar/20
let f(a)=∫_0 ^(π/4) (dx/(a +cosx +3sinx)) ⇒ f^′ (a)=−∫_0 ^(π/4) (dx/((a +cosx +3sinx)^2 )) ⇒ ∫_0 ^(π/4) (dx/((cosx +3sinx)^2 ))=−f^′ (0) let explicit f(a) f(a) =_(tan((x/2))=t) ∫_0 ^((√2)−1) (1/(a+((1−t^2 )/(1+t^2 ))+3((2t)/(1+t^2 ))))((2dt)/(1+t^2 )) = ∫_0 ^((√2)−1) ((2dt)/(a(1+t^2 )+1−t^2 +6t)) =∫_0 ^((√2)−1) (dt/((a−1)t^2 +6t +a+1)) Δ^′ =9−(a−1)(a+1) =9−(a^2 −1)=10−a^2 t_1 =((−3+(√(10−a^2 )))/(a−1)) and t_2 =((−3−(√(10−a^2 )))/(a−1)) (a≠1) f(a) =∫_0 ^((√2)−1) ((2dt)/((a−1)(t−t_1 )(t−t_2 ))) =(2/(a−1)) (1/(t_1 −t_2 ))∫_0 ^((√2)−1) {(1/(t−t_1 ))−(1/(t−t_2 ))}dt =(2/(a−1))×(1/((2(√(10−a^2 )))/(a−1)))[ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^((√2)−1) =(1/( (√(10−a^2 )))){ln∣(((√2)−1−t_1 )/( (√2)−1−t_2 ))∣−ln∣(t_1 /t_2 )∣} =(1/( (√(10−a^2 )))){ln∣(((√2)−1−((−3+(√(10−a^2 )))/(a−1)))/( (√2)−1+((3+(√(10−a^2 )))/(a−1))))∣ −ln∣((−3+(√(10−a^2 )))/(−3−(√(10−a^2 ))))∣
letf(a)=∫0π4dxa+cosx+3sinx⇒f′(a)=−∫0π4dx(a+cosx+3sinx)2⇒∫0π4dx(cosx+3sinx)2=−f′(0)letexplicitf(a)f(a)=tan(x2)=t∫02−11a+1−t21+t2+32t1+t22dt1+t2=∫02−12dta(1+t2)+1−t2+6t=∫02−1dt(a−1)t2+6t+a+1Δ′=9−(a−1)(a+1)=9−(a2−1)=10−a2t1=−3+10−a2a−1andt2=−3−10−a2a−1(a≠1)f(a)=∫02−12dt(a−1)(t−t1)(t−t2)=2a−11t1−t2∫02−1{1t−t1−1t−t2}dt=2a−1×1210−a2a−1[ln∣t−t1t−t2∣]02−1=110−a2{ln∣2−1−t12−1−t2∣−ln∣t1t2∣}=110−a2{ln∣2−1−−3+10−a2a−12−1+3+10−a2a−1∣−ln∣−3+10−a2−3−10−a2∣
Commented by abdomathmax last updated on 14/Mar/20
f(a) =(1/( (√(10−a^2 )))){ln∣((((√2)−1)(a−1)+3−(√(10−a^2 )))/(((√2)−1)(a−1)+3+(√(10−a^2 ))))∣ −ln∣((3−(√(10−a^2 )))/(3+(√(10−a^2 ))))∣ rest calculus of f^′ (a) and f^′ (0) ...
f(a)=110−a2{ln∣(2−1)(a−1)+3−10−a2(2−1)(a−1)+3+10−a2∣−ln∣3−10−a23+10−a2∣restcalculusoff′(a)andf′(0)…

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