Question Number 103537 by TMSF last updated on 15/Jul/20

∫(x/((a^2 cosx+b^2 sinx)))dx

Answered by mathmax by abdo last updated on 15/Jul/20

let I =∫  ((xdx)/(a^2 cosx +b^2 sinx))  we put a^2  =α and b^2  =β ⇒  I =∫  ((xdx)/(αcosx +βsinx)) changement tan((x/2))=t give  I =∫ ((2arctan(t))/(α×((1−t^2 )/(1+t^2 )) +β×((2t)/(1+t^2 ))))×((2dt)/(1+t^2 ))  =4∫  ((arctan(t))/(α−αt^2  +2βt))dt =−4 ∫  ((arctan(t))/(αt^2 −2βt −α))dt  let f(ξ) =∫  ((arctan(ξt))/(αt^2 −2βt −α))dt we have   f^′ (ξ) =∫  (t/((1+ξ^2 t^2 )(αt^2 −2βt −α)))dt  =_(ξt =u)    ∫  (u/(ξ(1+u^2 )(α(u^2 /ξ^2 )−2β(u/ξ)−α)))(du/ξ)  =∫  ((udu)/(ξ^2 (u^2 +1)(α (u^2 /ξ^2 ) −2β (u/ξ)−α))) =∫  ((udu)/((u^2  +1)(αu^2 −2βξu −αξ^2 )))  let decompose F(u) =(u/((u^2  +1)(αu^2 −2βξu −αξ^2 )))  αu^2  −2βξu −αξ^2  →Δ^′  =β^2 ξ^2 −α^2 ξ^2  =ξ^2 (β^2 −α^2 ) =ξ^2 (b^4 −a^4 )  case1  ∣b∣>∣a∣ ⇒u_1 =((βξ+ξ(√(β^2 −α^2 )))/α) and u_2 =((βξ−ξ(√(β^2 −α^2 )))/α) ⇒  F(u) =(u/((u^2  +1)α(u−u_1 )(u−u_2 ))) =(a_1 /(u−u_1 )) +(a_2 /(u−u_2 )) +((b_1 u +b_2 )/(u^2  +1))  eazy to find a_i  and b_i  ⇒  f^′ (ξ) =a_1 ln∣u−u_1 ∣ +a_2 ln∣u−u_2 ∣ +(b_1 /2)ln(u^2  +1)+b_2 arctan(u) +c ⇒  f(ξ) =a_1 ∫ ln∣u−u_1 ∣dξ +a_2 ∫ ln∣u−u_2 ∣dξ +(b_1 /2) ∫ ln∣u^2  +1∣dξ +b_2 ∫ arctanu dξ +cξ continued...