Question Number 78273 by msup trace by abdo last updated on 15/Jan/20
$${let}\:{f}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dx}}{\mathrm{1}+{sin}\theta\:{sinx}} \\ $$$$ \\ $$$${with}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$$$\left.\mathrm{1}\right)\:{explicite}\:{f}\left(\theta\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dx}}{\left(\mathrm{1}+{sin}\theta\:{sinx}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 18/Jan/20
![1) changement tan((x/2))=u ⇒f(θ)=∫_0 ^((√2)−1) (1/(1+sinθ×((2u)/(1+u^2 ))))((2du)/(1+u^2 )) =∫_0 ^((√2)−1) ((2du)/(1+u^2 +2sinθ u)) =∫_0 ^((√2)−1) ((2du)/(u^2 +2sinθ u +sin^2 θ +cos^2 θ)) =∫_0 ^((√2)−1) ((2du)/((u+sinθ)^2 +cos^2 θ)) =_(u+sinθ =(cosθ)t) 2 ∫_(tanθ) ^((((√2)−1)/(cosθ))+tanθ) ((cosθ dt)/(cos^2 θ(1+t^2 ))) =(2/(cosθ))[arctant]_(tanθ) ^((((√2)−1)/(cosθ))+tanθ) ⇒f(θ) =(2/(cosθ)){ arctan((((√2)−1)/(cosθ))+tanθ)−θ}](https://www.tinkutara.com/question/Q78582.png)
$$\left.\mathrm{1}\right)\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}\:\Rightarrow{f}\left(\theta\right)=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+{sin}\theta×\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{2}{sin}\theta\:{u}}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}{sin}\theta\:{u}\:+{sin}^{\mathrm{2}} \theta\:+{cos}^{\mathrm{2}} \theta} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{2}{du}}{\left({u}+{sin}\theta\right)^{\mathrm{2}} \:+{cos}^{\mathrm{2}} \theta}\:=_{{u}+{sin}\theta\:=\left({cos}\theta\right){t}} \mathrm{2}\:\int_{{tan}\theta} ^{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{{cos}\theta}+{tan}\theta} \:\:\frac{{cos}\theta\:{dt}}{{cos}^{\mathrm{2}} \theta\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{{cos}\theta}\left[{arctant}\right]_{{tan}\theta} ^{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{{cos}\theta}+{tan}\theta} \Rightarrow{f}\left(\theta\right)\:=\frac{\mathrm{2}}{{cos}\theta}\left\{\:{arctan}\left(\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{{cos}\theta}+{tan}\theta\right)−\theta\right\} \\ $$