Question Number 151221 by peter frank last updated on 19/Aug/21
$$\int\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\left(\mathrm{x}−\mathrm{a}\right)}\mathrm{dx} \\ $$
Commented by puissant last updated on 19/Aug/21
$${Q}=\int\frac{{sinx}}{{sin}\left({x}−{a}\right)}{dx} \\ $$$$=\int\frac{{sin}\left({x}−{a}+{a}\right)}{{sin}\left({x}−{a}\right)}{dx}\:\: \\ $$$$=\int\frac{{sin}\left({x}−{a}\right){cosa}+{cos}\left({x}−{a}\right){sina}}{{sin}\left({x}−{a}\right)}{dx} \\ $$$$=\int\left\{{cosa}+{cotan}\left({x}−{a}\right){sina}\right\}{dx} \\ $$$$ \\ $$$$\because\therefore\:\:{Q}\:=\:{xcosa}+{ln}\mid{sin}\left({x}−{a}\right)\mid{sina}+{C}.. \\ $$
Commented by peter frank last updated on 19/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Olaf_Thorendsen last updated on 19/Aug/21
![Let F_b (x) = ∫_b ^x ((sint)/(sin(t−a))) dt F_b (x) = ∫_(b−a) ^(x−a) ((sin(u+a))/(sinu)) du F_b (x) = ∫_(b−a) ^(x−a) ((sinucosa+cosusina)/(sinu)) du F_b (x) = [ucosa+ln∣sinu∣.sina]_(b−a) ^(x−a) F_b (x) = (x−b)cosa+ln∣((sin(x−a))/(sin(b−a)))∣.sina F(x) = xcosa+ln∣sin(x−a)∣+C](https://www.tinkutara.com/question/Q151228.png)
$$\mathrm{Let}\:\mathrm{F}_{{b}} \left({x}\right)\:=\:\int_{{b}} ^{{x}} \frac{\mathrm{sin}{t}}{\mathrm{sin}\left({t}−{a}\right)}\:{dt} \\ $$$$\mathrm{F}_{{b}} \left({x}\right)\:=\:\int_{{b}−{a}} ^{{x}−{a}} \frac{\mathrm{sin}\left({u}+{a}\right)}{\mathrm{sin}{u}}\:{du} \\ $$$$\mathrm{F}_{{b}} \left({x}\right)\:=\:\int_{{b}−{a}} ^{{x}−{a}} \frac{\mathrm{sin}{u}\mathrm{cos}{a}+\mathrm{cos}{u}\mathrm{sin}{a}}{\mathrm{sin}{u}}\:{du} \\ $$$$\mathrm{F}_{{b}} \left({x}\right)\:=\:\left[{u}\mathrm{cos}{a}+\mathrm{ln}\mid\mathrm{sin}{u}\mid.\mathrm{sin}{a}\right]_{{b}−{a}} ^{{x}−{a}} \\ $$$$\mathrm{F}_{{b}} \left({x}\right)\:=\:\left({x}−{b}\right)\mathrm{cos}{a}+\mathrm{ln}\mid\frac{\mathrm{sin}\left({x}−{a}\right)}{\mathrm{sin}\left({b}−{a}\right)}\mid.\mathrm{sin}{a} \\ $$$$\mathrm{F}\left({x}\right)\:=\:{x}\mathrm{cos}{a}+\mathrm{ln}\mid\mathrm{sin}\left({x}−{a}\right)\mid+\mathrm{C} \\ $$