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sin-x-sin-2-x-1-dx-

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

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