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Question Number 103537 by TMSF last updated on 15/Jul/20
∫x(a2cosx+b2sinx)dx
Answered by mathmax by abdo last updated on 15/Jul/20
letI=∫xdxa2cosx+b2sinxweputa2=αandb2=β⇒I=∫xdxαcosx+βsinxchangementtan(x2)=tgiveI=∫2arctan(t)α×1−t21+t2+β×2t1+t2×2dt1+t2=4∫arctan(t)α−αt2+2βtdt=−4∫arctan(t)αt2−2βt−αdtletf(ξ)=∫arctan(ξt)αt2−2βt−αdtwehavef′(ξ)=∫t(1+ξ2t2)(αt2−2βt−α)dt=ξt=u∫uξ(1+u2)(αu2ξ2−2βuξ−α)duξ=∫uduξ2(u2+1)(αu2ξ2−2βuξ−α)=∫udu(u2+1)(αu2−2βξu−αξ2)letdecomposeF(u)=u(u2+1)(αu2−2βξu−αξ2)αu2−2βξu−αξ2→Δ′=β2ξ2−α2ξ2=ξ2(β2−α2)=ξ2(b4−a4)case1∣b∣>∣a∣⇒u1=βξ+ξβ2−α2αandu2=βξ−ξβ2−α2α⇒F(u)=u(u2+1)α(u−u1)(u−u2)=a1u−u1+a2u−u2+b1u+b2u2+1eazytofindaiandbi⇒f′(ξ)=a1ln∣u−u1∣+a2ln∣u−u2∣+b12ln(u2+1)+b2arctan(u)+c⇒f(ξ)=a1∫ln∣u−u1∣dξ+a2∫ln∣u−u2∣dξ+b12∫ln∣u2+1∣dξ+b2∫arctanudξ+cξ...becontinued...
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