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Question Number 94914 by 174 last updated on 21/May/20

Answered by mathmax by abdo last updated on 22/May/20

A_+ =∫ x^2 (a^2 +x^2 )^(m−1)  dx     (m integr) ⇒x=at  A_+ =∫  a^2 t^2  a^(2m−2)  (1+t^2 )^(m−1) adt   =a^(2m+1)  ∫  t^2  Σ_(k=0) ^(m−1)  C_(m−1) ^k  t^(2k)  dt  =a^(2m+1)  Σ_(k=0) ^(m−1)  C_(m−1) ^k  t^(2k+2)  dt   =a^(2m+1)  Σ_(k=0) ^(m−1)   (C_(m−1) ^k /(2k+3)) t^(2k+3)   + C  =a^(3m+1)  Σ_(k=0) ^(m−1)  (C_(m+1) ^k /(2k+3))((x/a))^(3k+3)  +C    (we suppose a≠0)  A_− =∫ x^2 (a^2 −x^2 )^(m−1)  dx we do the changement x =asinα  ⇒A_− =∫ a^2  sin^2 α a^(2m−2)  (cos^2 α)^(m−1)  a cosα dα  =a^(2m+1)  ∫  sin^2 α  cos^(2m−1) α dα  =a^(2m+1 )  ∫ (1−cos^2 α)cos^(2m−1) α dα  =a^(2m+1)  ∫ cos^(2m−1) α dα −a^(2m+1)  ∫  cos^(2m+1)  α dα (wallis integral)  ∫ cos^(2m−1)  α dα =∫ (((e^(iα)  +e^(−iα) )/2))^(2m−1) dα  =(1/2^(2m−1) ) ∫  Σ_(k=0) ^(2m−1)  C_(2m−1) ^k  e^(ikα)  ×e^(−i(2m−1−k)α)  dα  =(1/2^(2m−1) ) Σ_(k=0) ^(2m−1)  C_(2m−1) ^k  ∫ e^(i(k−2m+1+k)α)  dα  =(1/2^(2m−1) ) Σ_(k=0) ^(2m−1)   C_(2m−1) ^k  ×(1/(2k−2m+1)) e^(i(2k−2m+1)α)  +C

A+=x2(a2+x2)m1dx(mintegr)x=atA+=a2t2a2m2(1+t2)m1adt=a2m+1t2k=0m1Cm1kt2kdt=a2m+1k=0m1Cm1kt2k+2dt=a2m+1k=0m1Cm1k2k+3t2k+3+C=a3m+1k=0m1Cm+1k2k+3(xa)3k+3+C(wesupposea0)A=x2(a2x2)m1dxwedothechangementx=asinαA=a2sin2αa2m2(cos2α)m1acosαdα=a2m+1sin2αcos2m1αdα=a2m+1(1cos2α)cos2m1αdα=a2m+1cos2m1αdαa2m+1cos2m+1αdα(wallisintegral)cos2m1αdα=(eiα+eiα2)2m1dα=122m1k=02m1C2m1keikα×ei(2m1k)αdα=122m1k=02m1C2m1kei(k2m+1+k)αdα=122m1k=02m1C2m1k×12k2m+1ei(2k2m+1)α+C

Commented by 174 last updated on 22/May/20

thanks a lot

Commented by ElOuafi last updated on 22/May/20

perfect sir !! thank you.

perfectsir!!thankyou.

Commented by mathmax by abdo last updated on 22/May/20

you are welcome

youarewelcome

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