Question Number 6710 by Tawakalitu. last updated on 15/Jul/16
$$\int\:\frac{{dx}}{{sinx}\:+\:{sin}\mathrm{2}{x}}\:{dx} \\ $$
Answered by Yozzii last updated on 15/Jul/16
$${I}=\int\frac{{dx}}{{sinx}+{sin}\mathrm{2}{x}}=\int\frac{{sinx}}{{sin}^{\mathrm{2}} {x}\left(\mathrm{1}+\mathrm{2}{cosx}\right)}{dx} \\ $$$${I}=\int\frac{−{sinx}}{−\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+\mathrm{2}{cosx}\right)}{dx} \\ $$$${Let}\:{u}={cosx}\Rightarrow{du}=−{sinxdx}. \\ $$$$\therefore{I}=\int\frac{{du}}{\left({u}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{1}+\mathrm{2}{u}\right)} \\ $$$${I}=\int\frac{\mathrm{1}}{\left({u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{2}{u}\right)}{du} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{u}−\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right)\frac{\mathrm{1}}{\mathrm{2}{u}+\mathrm{1}}{du} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\left\{\frac{\mathrm{1}}{\left({u}−\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({u}+\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)}\right\}{du} \\ $$$$\frac{\mathrm{0}.\mathrm{5}}{{u}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{u}+\mathrm{1}}=\frac{{u}+\mathrm{0}.\mathrm{5}−{u}+\mathrm{1}}{\left({u}−\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)}=\frac{\mathrm{3}/\mathrm{2}}{\left({u}−\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}\left({u}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}{u}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\left({u}−\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)} \\ $$$$ \\ $$$$\frac{\mathrm{0}.\mathrm{5}}{{u}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{u}+\mathrm{1}}=\frac{{u}+\mathrm{0}.\mathrm{5}−{u}−\mathrm{1}}{\left({u}+\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)}=\frac{−\mathrm{0}.\mathrm{5}}{\left({u}+\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\left({u}+\mathrm{1}\right)\left(\mathrm{2}{u}+\mathrm{1}\right)}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}{u}+\mathrm{1}}−\frac{\mathrm{0}.\mathrm{5}}{{u}+\mathrm{1}}\right)=\frac{\mathrm{2}}{\mathrm{2}{u}+\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}} \\ $$$$\therefore\:{I}=\frac{\mathrm{1}}{\mathrm{2}}\int\left\{\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}\left({u}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}{u}+\mathrm{1}}\right)−\left(\frac{\mathrm{2}}{\mathrm{2}{u}+\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right)\right\}{du} \\ $$$$ \\ $$$${I}=\int\left\{\frac{\mathrm{1}}{\mathrm{6}\left({u}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{2}{u}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}{u}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}\left({u}+\mathrm{1}\right)}\right\}{du} \\ $$$${I}=\int\left\{\frac{\mathrm{1}}{\mathrm{6}\left({u}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({u}+\mathrm{1}\right)}−\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{2}{u}+\mathrm{1}\right)}\right\}{du} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{6}}{ln}\mid{u}−\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{u}+\mathrm{1}\mid−\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid\mathrm{2}{u}+\mathrm{1}\mid+{C} \\ $$$$\int\frac{\mathrm{1}}{{sinx}+{sin}\mathrm{2}{x}}{dx}=\frac{\mathrm{1}}{\mathrm{6}}{ln}\mid{cosx}−\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{cosx}+\mathrm{1}\mid−\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid\mathrm{2}{cosx}+\mathrm{1}\mid+{C} \\ $$
Commented by Tawakalitu. last updated on 15/Jul/16
$${Wow}\:{thanks}\:{so}\:{much}. \\ $$