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Question Number 107238 by bemath last updated on 09/Aug/20

⊞bemath⊞ ∫ ((sin x)/(sin^3 x+cos^3 x)) dx ?

⊞bemath⊞∫sinxsin3x+cos3xdx?

Answered by john santu last updated on 09/Aug/20

⋇JS⋇ ∫ ((csc^2 x )/(1+cot^3 x)) dx = −∫ ((d(cot x))/(1+cot^3 x)) =−∫(dv/(1+v^3 )) [ with v = cot x ] = −∫ ((1−v^2 )/(1+v^3 )) dv −(1/3)∫ ((d(1+v^3 ))/(1+v^3 )) =−(1/2)∫ ((1+(1−2v))/(1+v^3 ))dv−(1/3)ln ∣1+v^3 ∣ =−(1/2)∫ (dv/((v−(1/2))^2 +(((√3)/2))^2 )) +(1/2)∫ ((d(1−v+v^2 ))/(1−v))−(1/3)ln ∣1+v^3 ∣ = −(1/(√3)) arc tan (((2cot x−1)/(√3)))+(1/6)ln ∣1−cot x+cot^2 x∣−(1/3)ln ∣1+cot x∣ + C

⋇JS⋇∫csc2x1+cot3xdx=−∫d(cotx)1+cot3x=−∫dv1+v3[withv=cotx]=−∫1−v21+v3dv−13∫d(1+v3)1+v3=−12∫1+(1−2v)1+v3dv−13ln∣1+v3∣=−12∫dv(v−12)2+(32)2+12∫d(1−v+v2)1−v−13ln∣1+v3∣=−13arctan(2cotx−13)+16ln∣1−cotx+cot2x∣−13ln∣1+cotx∣+C

Answered by som(math1967) last updated on 09/Aug/20

∫((sinxsec^3 x)/(sec^3 x(sin^3 x+cos^3 x)))dx ∫((tanx.sec^2 x)/(tan^3 x+1))dx let tanx=z ∴sec^2 xdx=dz ∫(z/(z^3 +1))dz =∫((z^2 +z−z^2 )/((z+1)(z^2 −z+1)))dz =∫(z/(z^2 −z+1))dz −∫(z^2 /(z^3 +1))dz =(1/2)∫((2z−1)/(z^2 −z+1))dz+(1/2)∫(dz/((z−(1/2))^2 +(((√3)/2))^2 )) −(1/3)∫((d(z^3 +1))/(z^3 +1)) =(1/2)∫((d(z^2 −z+1))/(z^2 −z+1)) +(1/2)∫(dz/((z−(1/2))^2 +(((√3)/2))^2 )) −(1/3)∫((d(z^3 +1))/(z^3 +1)) =(1/2)ln∣z^2 −z+1∣+(1/2)×(2/(√3))tan^(−1) ((z−(1/2))/((√3)/2)) −(1/3)ln∣z^3 +1∣+C now put z=tanx

∫sinxsec3xsec3x(sin3x+cos3x)dx∫tanx.sec2xtan3x+1dxlettanx=z∴sec2xdx=dz∫zz3+1dz=∫z2+z−z2(z+1)(z2−z+1)dz=∫zz2−z+1dz−∫z2z3+1dz=12∫2z−1z2−z+1dz+12∫dz(z−12)2+(32)2−13∫d(z3+1)z3+1=12∫d(z2−z+1)z2−z+1+12∫dz(z−12)2+(32)2−13∫d(z3+1)z3+1=12ln∣z2−z+1∣+12×23tan−1z−1232−13ln∣z3+1∣+Cnowputz=tanx

Answered by abdomathmax last updated on 09/Aug/20

I =∫ ((sinx)/(sin^3 x +cos^3 x))dx ⇒I =∫ ((sinx dx)/((sinx +cosx)(sin^2 x−cosxsinx +cos^2 x))) =∫ ((sinx dx)/((sinx +cosx)(1−cosx sinx))) =∫ (dx/((1+(1/(tanx)))(1−(1/2)sin(2x)))) =∫ ((tanx dx)/((1+tanx)(1−(1/2)×((2tanx)/(1+tan^2 x))))) =_(tanx =t) ∫ (t/((1+t)(1−(t/(1+t^2 )))))×(dt/(1+t^2 )) =∫ ((tdt)/((t+1)(1+t^2 −t))) =∫ ((tdt)/((t+1)(t^2 −t +1))) let decompose F(t) =(t/((t+1)(t^2 −t+1))) ⇒ F(t) =(a/(t+1)) +((bt +c)/(t^2 −t +1)) a =((−1)/3) , lim_(t→+∞) tF(t) =0 =a+b ⇒b =(1/3) ⇒ F(t)=((−1)/(3(t+1))) +(((t/3)+c)/(t^2 −t +1)) F(0) =0 =−(1/3) +c ⇒c =(1/3) ⇒ F(t) =−(1/(3(t+1))) +(1/3)((t+1)/(t^2 −t +1)) ⇒ ∫ F(t)dt =((−1)/3)ln∣t+1∣+(1/6)∫ ((2t−1+3)/(t^2 −t+1))dt =−(1/3)ln∣t+1∣+(1/6)ln(t^2 −t+1)+(1/2)∫(dt/(t^2 −t+1)) ∫ (dt/(t^2 −t+1)) =∫ (dt/((t−(1/2))^2 +(3/4))) =_(t−(1/2)=((√3)/2)u) =(4/3)∫ (1/(u^2 +1))((√3)/2)du =(2/(√3)) arctan(((2t−1)/(√3))) +C ⇒ I =−(1/3)ln∣t+1∣+(1/6)ln(t^2 −t+1)+(1/(√3)) arctan(((2t−1)/(√3))) +C =−(1/3)ln∣1+tanx∣+(1/6)ln(tan^2 x−tanx +1) +(1/(√3)) arctan(((2tanx−1)/(√3)))+C

I=∫sinxsin3x+cos3xdx⇒I=∫sinxdx(sinx+cosx)(sin2x−cosxsinx+cos2x)=∫sinxdx(sinx+cosx)(1−cosxsinx)=∫dx(1+1tanx)(1−12sin(2x))=∫tanxdx(1+tanx)(1−12×2tanx1+tan2x)=tanx=t∫t(1+t)(1−t1+t2)×dt1+t2=∫tdt(t+1)(1+t2−t)=∫tdt(t+1)(t2−t+1)letdecomposeF(t)=t(t+1)(t2−t+1)⇒F(t)=at+1+bt+ct2−t+1a=−13,limt→+∞tF(t)=0=a+b⇒b=13⇒F(t)=−13(t+1)+t3+ct2−t+1F(0)=0=−13+c⇒c=13⇒F(t)=−13(t+1)+13t+1t2−t+1⇒∫F(t)dt=−13ln∣t+1∣+16∫2t−1+3t2−t+1dt=−13ln∣t+1∣+16ln(t2−t+1)+12∫dtt2−t+1∫dtt2−t+1=∫dt(t−12)2+34=t−12=32u=43∫1u2+132du=23arctan(2t−13)+C⇒I=−13ln∣t+1∣+16ln(t2−t+1)+13arctan(2t−13)+C=−13ln∣1+tanx∣+16ln(tan2x−tanx+1)+13arctan(2tanx−13)+C

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