Question Number 129564 by bramlexs22 last updated on 16/Jan/21
$$\:\mathcal{V}\:=\:\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\left(\mathrm{x}+\theta\right)}\:\mathrm{dx}\: \\ $$
Answered by Lordose last updated on 16/Jan/21
$$\Omega\:=\:\int\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{x}+\theta\right)}\mathrm{dx}\:\overset{\mathrm{u}=\mathrm{x}+\theta} {=}\int\frac{\mathrm{sin}\left(\mathrm{u}−\theta\right)}{\mathrm{sin}\left(\mathrm{u}\right)}\mathrm{du} \\ $$$$\Omega\:=\:\int\frac{\mathrm{sin}\left(\mathrm{u}\right)\mathrm{cos}\theta−\mathrm{cos}\left(\mathrm{u}\right)\mathrm{sin}\theta}{\mathrm{sin}\left(\mathrm{u}\right)}\mathrm{du}\:=\:\mathrm{ucos}\theta\:−\:\mathrm{sin}\theta\mathrm{ln}\left(\mathrm{sinu}\right)\:+\:\mathrm{C} \\ $$$$\Omega\:=\:\left(\mathrm{x}+\theta\right)\mathrm{cos}\theta\:−\:\mathrm{sin}\theta\mathrm{ln}\left(\mathrm{sin}\left(\mathrm{x}+\theta\right)\right)+\:\mathrm{C} \\ $$
Answered by liberty last updated on 16/Jan/21
$$\mathcal{V}\:=\:\int\:\frac{\mathrm{sin}\:\left(\left(\mathrm{x}+\theta\right)−\theta\right)}{\mathrm{sin}\:\left(\mathrm{x}+\theta\right)}\mathrm{dx} \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{sin}\:\left(\mathrm{x}+\theta\right)\mathrm{cos}\:\theta−\mathrm{cos}\:\left(\mathrm{x}+\theta\right)\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\mathrm{x}+\theta\right)}\mathrm{dx} \\ $$$$\:\:\:=\int\:\mathrm{cos}\:\theta\:\mathrm{dx}−\mathrm{sin}\:\theta\int\frac{\mathrm{cos}\:\left(\mathrm{x}+\theta\right)}{\mathrm{sin}\:\left(\mathrm{x}+\theta\right)}\mathrm{dx} \\ $$$$\:\:\:=\:\mathrm{xcos}\:\theta−\mathrm{sin}\:\theta.\mathrm{ln}\:\mid\mathrm{sin}\:\left(\mathrm{x}+\theta\right)\mid\:+\:\mathrm{C} \\ $$