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Question Number 95562 by peter frank last updated on 26/May/20
show that  ∫((sin (x−θ))/(sin x))dx=xcos x−sin θlog sin x
$$\mathrm{show}\:\mathrm{that} \\ $$$$\int\frac{\mathrm{sin}\:\left(\mathrm{x}−\theta\right)}{\mathrm{sin}\:\mathrm{x}}\mathrm{dx}=\mathrm{xcos}\:\mathrm{x}−\mathrm{sin}\:\theta\mathrm{log}\:\mathrm{sin}\:\mathrm{x} \\ $$
Answered by Ar Brandon last updated on 26/May/20
sin(x−𝛉)=sin(x)cos(𝛉)−cos(x)sin(𝛉)  ⇒I=∫((sin(x)cos(𝛉)−cos(x)sin(𝛉))/(sin(x)))dx=∫cos(𝛉)dx−∫((cos(x))/(sin(x)))sin(𝛉)d𝛉         =xcos(𝛉)−sin(𝛉)log(sin(x))+C
$$\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\theta}\right)=\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\theta}\right)−\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right) \\ $$$$\Rightarrow\mathrm{I}=\int\frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\theta}\right)−\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)}{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}\boldsymbol{\mathrm{dx}}=\int\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\theta}\right)\boldsymbol{\mathrm{dx}}−\int\frac{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)\boldsymbol{\mathrm{d}\theta} \\ $$$$\:\:\:\:\:\:\:=\boldsymbol{\mathrm{xcos}}\left(\boldsymbol{\theta}\right)−\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\right)+\boldsymbol{\mathrm{C}} \\ $$

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