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Question Number 54701 by Meritguide1234 last updated on 09/Feb/19

Answered by behi83417@gmail.com last updated on 09/Feb/19

tg^(−1) (x/(x^2 −1))=a  sin^2 a=(1/(1+cot^2 a))=(1/(1+x^2 +x^(−2) −2))=(x^2 /(x^4 −x^2 +1))  ⇒(√(x^4 −x^2 +1))=(x/(sina))  cos^2 a=(1/(1+tg^2 a))=(1/(1+(x^2 /((x^2 −1)^2 ))))=(((x^2 −1)^2 )/(x^4 −x^2 +1))  ⇒cosa=((x^2 −1)/(√(x^4 −x^2 +1)))  sin(x+a)=sinx.((x^2 −1)/(√(x^4 −x^2 +1)))+cosx.(x/(√(x^4 −x^2 +1)))  tg^(−1) ((x^2 −1)/(2x))=b⇒cos^2 b=(1/(1+(((x^2 −1)^2 )/(4x^2 ))))⇒  cosb=((2x)/(x^2 +1)),sinb=((x^2 −1)/(x^2 +1))  sin(2x−b)=sin2x.((2x)/(x^2 +1))−cos2x.((x^2 −1)/(x^2 +1))  ⇒I=∫(((x^2 −1).sinx+x.cosx)/([(x^2 +1)+(x^2 −1)cos2x−2xsin2x]))dx

tg1xx21=asin2a=11+cot2a=11+x2+x22=x2x4x2+1x4x2+1=xsinacos2a=11+tg2a=11+x2(x21)2=(x21)2x4x2+1cosa=x21x4x2+1sin(x+a)=sinx.x21x4x2+1+cosx.xx4x2+1tg1x212x=bcos2b=11+(x21)24x2cosb=2xx2+1,sinb=x21x2+1sin(2xb)=sin2x.2xx2+1cos2x.x21x2+1I=(x21).sinx+x.cosx[(x2+1)+(x21)cos2x2xsin2x]dx

Commented by Meritguide1234 last updated on 09/Feb/19

check again

checkagain

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

  tan^(−1) ((x/(x^2 −1)))  sec^2 a=((x^4 −2x^2 +1+x^2 )/((x^2 −1)^2 ))=((x^4 −x^2 +1)/((x^2 −1)^2 ))  cos^2 a=(((x^2 −1)^2 )/(x^4 −x^2 +1)) cosa=((x^2 −1)/(√(x^4 −x^2 +1)))  sin^2 a=1−(((x^4 −2x^2 +1))/(x^4 −x^2 +1))=((x^4 −x^2 +1−x^4 +2x^2 −1)/(x^4 −x^2 +1))  sina=(x/(√(x^4 −x^2 +1)))  so sin(x+tan^(−1) (x/(x^2 −1)))=sinx×((x^2 −1)/(√(x^4 −x^2 +1)))+cosx×(x/(√(x^4 −x^2 +1)))  so N_r ={(x^2 −1)sinx+xcosx}×(1/(√(x^4 −x^2 +1)))×(√(x^4 −x^2 +1))  N_r =(x^2 −1)sinx+xcosx  ★tanb=((x^2 −1)/(2x))  sec^2 b=1+((x^4 −2x^2 +1)/(4x^2 ))=(((x^2 +1)^2 )/((2x)^2 ))  cosb=((2x)/((x^2 +1)))  sin^2 b=(((x^2 +1)^2 −4x^2 )/((x^2 +1)^2 ))=(((x^2 −1)/(x^2 +1)))^2   sinb=((x^2 −1)/(x^2 +1)) or ((1−x^2 )/(1+x^2 ))  sin(2x−tan^(−1) ((x^2 −1)/(2x)))  =sin2x×((2x)/(x^2 +1))−co2x×((x^2 −1)/(x^2 +1))  or sin2x×((2x)/(x^2 +1))−cos2x×((1−x^2 )/(1+x^2 ))  ((2xsin2x−(x^2 −1)cos2x)/(x^2 +1))  =(((2xsin2x−(x^2 −1)cos2x)/(x^2 +1))  ((2xsin2x−(1−x^2 )cos2x)/(x^2 +1))  so D_r =(x^2 +1){1−((2xsin2x−(x^2 −1)cos2x)/(x^2 +1))}  =x^2 +1−2xsin2x+(x^2 −1)cos2x  =x^2 +1−4xsinxcosx+x^2 cos2x−cos2x  =x^2 (1+cos2x)+1−cos2x−4xsinxcosx  =2x^2 cos^2 x+2sin^2 x−4xsinxcosx  =2(xcosx−sinx)^2  or2(sinx−xcosx)^2     or D_r =(x^2 +1){1−((2xsin2x−(1−x^2 )cos2x)/(x^2 +1))}  =x^2 +1−2xsin2x+(1−x^2 )cos2x  =x^2 +1−4xsinxcosx+cos2x−x^2 cos2x  =x^2 (1−cos2x)+1+cos2x−4xsinxcosx  =2x^2 sin^2 x+2cos^2 x−4xsinxcosx  =2(xsinx−cosx)^2  or2(cosx−xsinx)^2   I_(blue) =∫(((x^2 −1)sinx+xcosx)/(2(xcosx−sinx)^2 ))dx.....1  or ∫(((x^2 −1)sinx+xcosx)/(2(sinx−xcosx)^2 ))dx.......2  I_(red) =∫(((x^2 −1)sinx+xcosx)/(2(xsinx−cosx)^2 ))dx.....3  or∫(((x^2 −1)sinx+xcosx)/(2(cosx−xsinx)^2 ))dx....4  four intregation...  (d/dx)(xcosx−sinx)=−xsinx+cosx−cosx=−xsinx  (d/dx)(sinx−xcosx)=cosx−(x×−sinx+cosx)=xsinx  (1/2)∫(/)  (d/(dx ))(xsinx−cosx)=xcosx+sinx+sinx=xcosx+2sinx  (d/dx)(cosx−xsinx)=−sinx−xcosx−sinx=−(xcosx+2sinx)

tan1(xx21)sec2a=x42x2+1+x2(x21)2=x4x2+1(x21)2cos2a=(x21)2x4x2+1cosa=x21x4x2+1sin2a=1(x42x2+1)x4x2+1=x4x2+1x4+2x21x4x2+1sina=xx4x2+1sosin(x+tan1xx21)=sinx×x21x4x2+1+cosx×xx4x2+1soNr={(x21)sinx+xcosx}×1x4x2+1×x4x2+1Nr=(x21)sinx+xcosxtanb=x212xsec2b=1+x42x2+14x2=(x2+1)2(2x)2cosb=2x(x2+1)sin2b=(x2+1)24x2(x2+1)2=(x21x2+1)2sinb=x21x2+1or1x21+x2sin(2xtan1x212x)=sin2x×2xx2+1co2x×x21x2+1orsin2x×2xx2+1cos2x×1x21+x22xsin2x(x21)cos2xx2+1=(2xsin2x(x21)cos2xx2+12xsin2x(1x2)cos2xx2+1soDr=(x2+1){12xsin2x(x21)cos2xx2+1}=x2+12xsin2x+(x21)cos2x=x2+14xsinxcosx+x2cos2xcos2x=x2(1+cos2x)+1cos2x4xsinxcosx=2x2cos2x+2sin2x4xsinxcosx=2(xcosxsinx)2or2(sinxxcosx)2orDr=(x2+1){12xsin2x(1x2)cos2xx2+1}=x2+12xsin2x+(1x2)cos2x=x2+14xsinxcosx+cos2xx2cos2x=x2(1cos2x)+1+cos2x4xsinxcosx=2x2sin2x+2cos2x4xsinxcosx=2(xsinxcosx)2or2(cosxxsinx)2Iblue=(x21)sinx+xcosx2(xcosxsinx)2dx.....1or(x21)sinx+xcosx2(sinxxcosx)2dx.......2Ired=(x21)sinx+xcosx2(xsinxcosx)2dx.....3or(x21)sinx+xcosx2(cosxxsinx)2dx....4fourintregation...ddx(xcosxsinx)=xsinx+cosxcosx=xsinxddx(sinxxcosx)=cosx(x×sinx+cosx)=xsinx12ddx(xsinxcosx)=xcosx+sinx+sinx=xcosx+2sinxddx(cosxxsinx)=sinxxcosxsinx=(xcosx+2sinx)

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Feb/19

i will try to solve in copy then upload it...

iwilltrytosolveincopythenuploadit...

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