Question Number 152115 by peter frank last updated on 25/Aug/21
$$\int\frac{\mathrm{dx}}{\mathrm{asin}\:\mathrm{x}+\mathrm{bcos}\:\mathrm{x}} \\ $$
Answered by aleks041103 last updated on 26/Aug/21
$$\int\frac{\mathrm{dx}}{\mathrm{asin}\:\mathrm{x}+\mathrm{bcos}\:\mathrm{x}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\int\frac{{dx}}{{sin}\left({x}+{r}\right)},\:{r}={atan}\left({b}/{a}\right) \\ $$$$\int\frac{{dx}}{{sin}\:{x}}=\int\frac{{sin}\:{x}\:{dx}}{\mathrm{1}−{cos}^{\mathrm{2}} {x}}=−\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} }= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}+\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\right)={ln}\sqrt{\frac{\mathrm{1}−{cos}\:{x}}{\mathrm{1}+{cos}\:{x}}}+{C} \\ $$$$\int\frac{{dx}}{{asin}\left({x}\right)+{bcos}\left({x}\right)}=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{ln}\left(\frac{\mathrm{1}−{cos}\left({x}+{atan}\left({b}/{a}\right)\right)}{\mathrm{1}+{cos}\left({x}+{atan}\left({b}/{a}\right)\right)}\right)+{C} \\ $$$$ \\ $$
Commented by peter frank last updated on 26/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$