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Question Number 32363 by prof Abdo imad last updated on 23/Mar/18
let consider the function f(x,θ) = ∫_x ^x^2 ln( 2+sinθ cost)dt calculate (∂f/∂x)(x,θ) and (∂f/∂θ)(x,θ) .
$${let}\:{consider}\:{the}\:{function} \\ $$$${f}\left({x},\theta\right)\:=\:\:\int_{{x}} ^{{x}^{\mathrm{2}} } {ln}\left(\:\mathrm{2}+{sin}\theta\:{cost}\right){dt} \\ $$$${calculate}\:\frac{\partial{f}}{\partial{x}}\left({x},\theta\right)\:{and}\:\:\frac{\partial{f}}{\partial\theta}\left({x},\theta\right)\:. \\ $$
Commented by prof Abdo imad last updated on 25/Mar/18
(∂f/∂x)(x,θ) = 2x ln(2 +sinθ cos(x^2 ))−ln(2+sinθ cosx) (∂f/∂θ)(x,θ) = ∫_x ^x^2 ((cost cosθ)/(2+costsinθ)) dt ch. tan((t/2))=ugive (∂f/∂θ) = ∫_(tan((x/2))) ^(tan((x^2 /2))) ((cosθ((1−u^2 )/(1+u^2 )))/(2+sinθ((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫_(tan((x/2))) ^(tan((x^2 /2))) (((1−u^2 )cosθ)/((1+u^2 )(2 +(1−u^2 )sinθ)))du = ∫_(tan((x/2))) ^(tan((x^2 /2))) ((α −αu^2 )/((1+u^2 )(2+β −βu^2 ))) du whit α=cosθ and β=sinθ =∫_(tan((x/2))) ^(tan((x^2 /2)) ) ((−αu^2 +α)/((1+u^2 )(−βu^2 +β+2)))du let decompose f(u) = ((αu^2 −α)/((1+u^2 )(βu^2 −β−2))) = ((αu^2 −α)/(β(1+u^2 )( u^2 −((√((β+2)/β)))^2 ))) = (a/((u−(√((β+2)/β))))) +(b/(u+(√((β+2)/β)))) + ((cu +d)/(u^2 +1)) ...be conyinued....
$$\frac{\partial{f}}{\partial{x}}\left({x},\theta\right)\:=\:\mathrm{2}{x}\:{ln}\left(\mathrm{2}\:+{sin}\theta\:{cos}\left({x}^{\mathrm{2}} \right)\right)−{ln}\left(\mathrm{2}+{sin}\theta\:{cosx}\right) \\ $$$$\frac{\partial{f}}{\partial\theta}\left({x},\theta\right)\:=\:\int_{{x}} ^{{x}^{\mathrm{2}} } \:\:\frac{{cost}\:{cos}\theta}{\mathrm{2}+{costsin}\theta}\:{dt}\:\:{ch}.\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={ugive} \\ $$$$\frac{\partial{f}}{\partial\theta}\:=\:\int_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} ^{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)} \:\:\:\frac{{cos}\theta\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{2}+{sin}\theta\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} ^{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)} \:\:\:\:\:\frac{\left(\mathrm{1}−{u}^{\mathrm{2}} \right){cos}\theta}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}\:+\left(\mathrm{1}−{u}^{\mathrm{2}} \right){sin}\theta\right)}{du} \\ $$$$=\:\int_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} ^{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)} \:\:\:\frac{\alpha\:−\alpha{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}+\beta\:−\beta{u}^{\mathrm{2}} \right)}\:{du}\:{whit}\:\alpha={cos}\theta \\ $$$${and}\:\beta={sin}\theta \\ $$$$=\int_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} ^{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\:\:} \:\:\:\:\:\frac{−\alpha{u}^{\mathrm{2}} \:+\alpha}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(−\beta{u}^{\mathrm{2}} +\beta+\mathrm{2}\right)}{du}\:{let}\:{decompose} \\ $$$${f}\left({u}\right)\:=\:\:\frac{\alpha{u}^{\mathrm{2}} \:−\alpha}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\beta{u}^{\mathrm{2}} −\beta−\mathrm{2}\right)} \\ $$$$=\:\:\frac{\alpha{u}^{\mathrm{2}} \:−\alpha}{\beta\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\:{u}^{\mathrm{2}} \:−\left(\sqrt{\frac{\beta+\mathrm{2}}{\beta}}\right)^{\mathrm{2}} \right)} \\ $$$$=\:\:\frac{{a}}{\left({u}−\sqrt{\frac{\beta+\mathrm{2}}{\beta}}\right)}\:\:+\frac{{b}}{{u}+\sqrt{\frac{\beta+\mathrm{2}}{\beta}}}\:\:+\:\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:…{be} \\ $$$${conyinued}…. \\ $$

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