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Question Number 41703 by abdo.msup.com last updated on 11/Aug/18
calculate I = ∫_0 ^(π/4) ((cosx)/(cos^3 x +sin^3 x))dx
calculateI=∫0π4cosxcos3x+sin3xdx
Commented by math khazana by abdo last updated on 12/Aug/18
I = ∫_0 ^(π/4) (((cosx)/(cos^3 x))/((cos^3 x +sin^3 x)/(cos^3 x))) dx =∫_0 ^(π/4) (1/(cos^2 x(1+tan^3 x)))dx =∫_0 ^(π/4) ((1+tan^2 x)/(1+tan^3 x)) dx =_(tanx=t) ∫_0 ^1 ((1+t^2 )/(1+t^3 )) (dt/(1+t^2 )) = ∫_0 ^1 (dt/(1+t^3 )) let decompose F(t)= (1/(t^3 +1)) =(1/((t+1)(t^2 −t +1))) F(t)=(a/(t+1)) +((bt +c)/(t^2 −t +1)) a =lim_(t→−1) (t+1)F(t) =(1/3) lim_(t→+∞) tF(t)=0=a +b ⇒b=−(1/3) ⇒ F(t) = (1/(3(t+1))) +((−(1/3)t +c)/(t^2 −t +1)) F(0) =1 =(1/3) +c ⇒c=(2/3) ⇒ F(t) = (1/(3(t+1))) −(1/3) ((t−2)/(t^2 −t +1)) ⇒ ∫_0 ^1 F(t)dt =(1/3)[ln∣t+1∣]_0 ^1 −(1/3) ∫_0 ^1 ((t−2)/(t^2 −t +1))dt =((ln(2))/3) −(1/6) ∫_0 ^1 ((2t−1−3)/(t^2 −t +1)) dt =((ln(2))/3) −[(1/6)ln∣t^2 −t+1∣ ]_0 ^1 +(1/2) ∫_0 ^1 (dt/(t^2 −t +1)) =((ln(2))/3) +(1/2) ∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) =_(t−(1/2)=((√3)/2)u) ((ln(2))/3) +(1/2) (4/3) ∫_(−(1/( (√3)))) ^(1/( (√3))) (1/(1+u^2 )) ((√3)/2) du =((ln(2))/3) + ((√3)/3) [ arctanu]_(−(1/( (√3)))) ^(1/( (√3))) =((ln(2))/3) + (1/( (√3))){ 2 arctan((1/( (√3))))} =((ln(2))/3) +(2/( (√3))) (π/6) ⇒ I = ((ln(2))/3) +(π/(3(√3))) .
I=∫0π4cosxcos3xcos3x+sin3xcos3xdx=∫0π41cos2x(1+tan3x)dx=∫0π41+tan2x1+tan3xdx=tanx=t∫011+t21+t3dt1+t2=∫01dt1+t3letdecomposeF(t)=1t3+1=1(t+1)(t2−t+1)F(t)=at+1+bt+ct2−t+1a=limt→−1(t+1)F(t)=13limt→+∞tF(t)=0=a+b⇒b=−13⇒F(t)=13(t+1)+−13t+ct2−t+1F(0)=1=13+c⇒c=23⇒F(t)=13(t+1)−13t−2t2−t+1⇒∫01F(t)dt=13[ln∣t+1∣]01−13∫01t−2t2−t+1dt=ln(2)3−16∫012t−1−3t2−t+1dt=ln(2)3−[16ln∣t2−t+1∣]01+12∫01dtt2−t+1=ln(2)3+12∫01dt(t−12)2+34=t−12=32uln(2)3+1243∫−131311+u232du=ln(2)3+33[arctanu]−1313=ln(2)3+13{2arctan(13)}=ln(2)3+23π6⇒I=ln(2)3+π33.

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