Question Number 63738 by MJS last updated on 08/Jul/19
![useful formula ======== ∀a∈R^+ :∀b ∈R: a sin x +b cos x =(√(a^2 +b^2 ))sin (x+arctan (b/a)) ∫(dx/(a sin x +b cos x))= =(1/( (√(a^2 +b^2 ))))∫(dx/(sin (x+arctan (b/a))))= [t=x+arctan (b/a) → dx=dt] (1/( (√(a^2 +b^2 ))))∫(dt/(sin t))=−(1/( (√(a^2 +b^2 ))))ln ((1/(sin t))+(1/(tan t))) = =−(1/( (√(a^2 +b^2 ))))ln ∣(((√(a^2 +b^2 ))−b sin x +a cos x)/(a sin x +b cos x))∣ +C](https://www.tinkutara.com/question/Q63738.png)
$$\mathrm{useful}\:\mathrm{formula} \\ $$$$======== \\ $$$$ \\ $$$$\forall{a}\in\mathbb{R}^{+} :\forall{b}\:\in\mathbb{R}:\:{a}\:\mathrm{sin}\:{x}\:+{b}\:\mathrm{cos}\:{x}\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{{b}}{{a}}\right) \\ $$$$\int\frac{{dx}}{{a}\:\mathrm{sin}\:{x}\:+{b}\:\mathrm{cos}\:{x}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\int\frac{{dx}}{\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{{b}}{{a}}\right)}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\mathrm{arctan}\:\frac{{b}}{{a}}\:\:\rightarrow\:{dx}={dt}\right] \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\int\frac{{dt}}{\mathrm{sin}\:{t}}=−\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:{t}}+\frac{\mathrm{1}}{\mathrm{tan}\:{t}}\right)\:= \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\mathrm{ln}\:\mid\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{b}\:\mathrm{sin}\:{x}\:+{a}\:\mathrm{cos}\:{x}}{{a}\:\mathrm{sin}\:{x}\:+{b}\:\mathrm{cos}\:{x}}\mid\:+{C} \\ $$