Question Number 152165 by peter frank last updated on 26/Aug/21
$$\int\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{3x}}\mathrm{dx} \\ $$
Answered by john_santu last updated on 26/Aug/21
$$\:\mathrm{I}=\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{x}}\:\mathrm{dx}\: \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{dx}}{\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{Apply}\:\mathrm{Wierstrass}\:\mathrm{subtitution}\: \\ $$$$\mathrm{with}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{u}\: \\ $$
Commented by peter frank last updated on 26/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by puissant last updated on 26/Aug/21
$$=\int\frac{{sinx}}{\mathrm{3}{sinx}−\mathrm{4}{sin}^{\mathrm{3}} {x}}{dx}=\int\frac{\mathrm{1}}{\mathrm{3}−\mathrm{4}{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int\frac{{sec}^{\mathrm{2}} {x}}{\mathrm{3}{sec}^{\mathrm{2}} {x}−\mathrm{4}{tan}^{\mathrm{2}} {x}}{dx}=\int\frac{{sec}^{\mathrm{2}} {x}}{\mathrm{3}−{tan}^{\mathrm{2}} }{dx} \\ $$$$\overset{{t}={tanx}} {=}−\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{3}} \\ $$$$\Rightarrow\:{I}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\mid\frac{{t}−\sqrt{\mathrm{3}}}{{t}+\sqrt{\mathrm{3}}}\mid+{C} \\ $$$$ \\ $$$$\:\:\:\therefore\because\:{I}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\mid\frac{{tanx}−\sqrt{\mathrm{3}}}{{tanx}+\sqrt{\mathrm{3}}}\mid+{C}.. \\ $$
Commented by peter frank last updated on 26/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$