Question Number 83521 by M±th+et£s last updated on 03/Mar/20
$$\int\frac{\sqrt{{sin}\left({x}\right)}}{{sin}^{\mathrm{2}} \left({x}\right)+\mathrm{1}}\:{dx} \\ $$
Commented by msup trace by abdo last updated on 04/Mar/20
$${changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}=\int\frac{\sqrt{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}}{\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{1}}×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{\frac{\sqrt{\mathrm{2}{t}}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}}{\frac{\mathrm{4}{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}+\mathrm{1}}×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int\:\:\:\:\frac{\sqrt{\mathrm{2}{t}}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\left(\frac{\mathrm{4}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\int\:\:\:\frac{\sqrt{\mathrm{2}{t}}}{\frac{\mathrm{4}{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}+\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt} \\ $$$$=\int\:\:\:\frac{\sqrt{\mathrm{2}{t}}\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)}{\mathrm{4}{t}^{\mathrm{2}} +\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=_{\sqrt{{t}}={u}} \:\:\:\:\sqrt{\mathrm{2}}\int\:\:\:\:\frac{{u}\sqrt{\mathrm{1}+{u}^{\mathrm{4}} }}{\mathrm{4}{u}^{\mathrm{4}} +\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\int\:\:\frac{{u}^{\mathrm{2}} \sqrt{\mathrm{1}+{u}^{\mathrm{4}} }}{\mathrm{4}{u}^{\mathrm{4}} +{u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}}{du} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\int\:\:\frac{{u}^{\mathrm{2}} \sqrt{\mathrm{1}+{u}^{\mathrm{4}} }}{\mathrm{5}{u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}}{du}….{be}\:{continued}… \\ $$