x-a-2-cosx-b-2-sinx-dx-

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

\int \frac{x}{(a^{2} cosx + b^{2} sinx)} dx

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

let I = \int \frac{xdx}{a^{2} cosx + b^{2} sinx} we put a^{2} = α and b^{2} = β \Rightarrow

I = \int \frac{xdx}{α cosx + β sinx} changement \tan (\frac{x}{2}) = t give

I = \int \frac{2 \arctan (t)}{α \times \frac{1 - t^{2}}{1 + t^{2}} + β \times \frac{2 t}{1 + t^{2}}} \times \frac{2 dt}{1 + t^{2}}

= 4 \int \frac{\arctan (t)}{α - α t^{2} + 2 β t} dt = - 4 \int \frac{\arctan (t)}{α t^{2} - 2 β t - α} dt

let f (ξ) = \int \frac{\arctan (ξ t)}{α t^{2} - 2 β t - α} dt we have

f^{^{'}} (ξ) = \int \frac{t}{(1 + ξ^{2} t^{2}) (α t^{2} - 2 β t - α)} dt

=_{ξ t = u} \int \frac{u}{ξ (1 + u^{2}) (α \frac{u^{2}}{ξ^{2}} - 2 β \frac{u}{ξ} - α)} \frac{du}{ξ}

= \int \frac{udu}{ξ^{2} (u^{2} + 1) (α \frac{u^{2}}{ξ^{2}} - 2 β \frac{u}{ξ} - α)} = \int \frac{udu}{(u^{2} + 1) (α u^{2} - 2 β ξ u - α ξ^{2})}

let decompose F (u) = \frac{u}{(u^{2} + 1) (α u^{2} - 2 β ξ u - α ξ^{2})}

α u^{2} - 2 β ξ u - α ξ^{2} \to Δ^{^{'}} = β^{2} ξ^{2} - α^{2} ξ^{2} = ξ^{2} (β^{2} - α^{2}) = ξ^{2} (b^{4} - a^{4})

case 1 ∣ b ∣>∣ a ∣ \Rightarrow u_{1} = \frac{β ξ + ξ \sqrt{β^{2} - α^{2}}}{α} and u_{2} = \frac{β ξ - ξ \sqrt{β^{2} - α^{2}}}{α} \Rightarrow

F (u) = \frac{u}{(u^{2} + 1) α (u - u_{1}) (u - u_{2})} = \frac{a_{1}}{u - u_{1}} + \frac{a_{2}}{u - u_{2}} + \frac{b_{1} u + b_{2}}{u^{2} + 1}

eazy to find a_{i} and b_{i} \Rightarrow

f^{^{'}} (ξ) = a_{1} \ln ∣ u - u_{1} ∣ + a_{2} \ln ∣ u - u_{2} ∣ + \frac{b_{1}}{2} \ln (u^{2} + 1) + b_{2} \arctan (u) + c \Rightarrow

f (ξ) = a_{1} \int \ln ∣ u - u_{1} ∣ d ξ + a_{2} \int \ln ∣ u - u_{2} ∣ d ξ + \frac{b_{1}}{2} \int \ln ∣ u^{2} + 1 ∣ d ξ + b_{2} \int arctanu d ξ + c ξ

\dots be continued \dots