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Question Number 85718 by M±th+et£s last updated on 24/Mar/20

∫((sin(x)−cos(3x))/(sin(x)−cos(2x)))dx

sin(x)cos(3x)sin(x)cos(2x)dx

Answered by MJS last updated on 24/Mar/20

∫((sin x −cos 3x)/(sin x −cos 2x))dx=  =∫((4cos x sin^2  x −cos x +sin x)/(−2cos^2  x +sin x +1))dx=       [t=tan (x/2) → dx=((2dt)/(t^2 +1))]  =−2∫((t^6 +2t^5 −15t^4 +4t^3 +15t^2 +2t−1)/(t^8 −2t^7 −4t^6 −6t^5 −10t^4 −6t^3 −4t^2 −2t+1))dt=  =−2∫((...)/((t^2 +1)^2 (t+1)^2 (t−2−(√3))(t−2+(√3))))dt=  =8∫(dt/((t^2 +1)^2 ))+∫((2t)/(t^2 +1))dt−4∫(dt/(t^2 +1))+(2/3)∫(dt/((t^2 +1)^2 ))−2∫(dt/(t+1))−((√3)/9)∫(dt/(t−2−(√3)))+((√3)/9)∫(dt/(t−2+(√3)))  now it′s easy

sinxcos3xsinxcos2xdx==4cosxsin2xcosx+sinx2cos2x+sinx+1dx=[t=tanx2dx=2dtt2+1]=2t6+2t515t4+4t3+15t2+2t1t82t74t66t510t46t34t22t+1dt==2...(t2+1)2(t+1)2(t23)(t2+3)dt==8dt(t2+1)2+2tt2+1dt4dtt2+1+23dt(t2+1)22dtt+139dtt23+39dtt2+3nowitseasy

Commented by MJS last updated on 24/Mar/20

I get  ((2(5t^2 +6t−1))/(3(t^2 +1)(t+1)))+ln ((t^2 +1)/((t+1)^2 )) +((√3)/9)ln ((t−2+(√3))/(t−2−(√3)))  with t=tan (x/2)

Iget2(5t2+6t1)3(t2+1)(t+1)+lnt2+1(t+1)2+39lnt2+3t23witht=tanx2

Commented by M±th+et£s last updated on 24/Mar/20

thank you sir . god bless you

thankyousir.godblessyou

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