Question Number 97400 by student work last updated on 07/Jun/20
$$\int\frac{\mathrm{xdx}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{3}}=? \\ $$$$\:\:\mathrm{help}\:\mathrm{me} \\ $$
Commented by Tony Lin last updated on 08/Jun/20
![sinx=((e^(ix) −e^(−ix) )/(2i)) ∫(x/(−(((e^(ix) −e^(−ix) )/2))^2 −3))dx =−4∫(x/((e^(ix) −e^(−ix) )^2 +12))dx =−4∫((xe^(2ix) )/((e^(2ix) −1)^2 +12e^(2ix) ))dx let u=e^(2ix) −1, dx=(du/(2ie^(2ix) )) x=−((iln(u+1))/2) −∫((ln(u+1))/(u^2 +12+12))du =−((ln(u+1))/([u−(2(√6)−6)][u−(−2(√6)−6]))du =(1/(4(√6)))[∫((ln(u+1))/(u+2(√6)+6))du−∫((ln(u+1))/(u−2(√6)+6))du] let v=u+2(√6)+6,dv=du let t=u−2(√6)+6,dt=du (1/(4(√6)))[∫((ln(v−2(√6)−5))/v)dv−∫((ln(t+2(√6)−5))/t)dt] =(1/(4(√6)))∫[(( ln((v/(−2(√6)−5))+1))/v)+ln(−2(√6)−5)(1/v)]dv −(1/(4(√6)))∫[((ln((t/(2(√6)−5))+1))/t)+ln(2(√6)−5)(1/v)]dv let w=−(v/(−2(√6)−5)) ,dv=(5+2(√6))dw let z=−(t/(−2(√6)+5)) ,dt=(5−2(√6))dz (1/(4(√6))){−∫−((ln(1−w))/w)dw+∫−((ln(1−z))/z)dz +ln(−2(√6)−5)ln∣v∣+ln(2(√6)−5)ln∣t∣} =(1/(4(√6)))[Li_2 (w)−Li_2 (z)+ln(−2(√6)−5)ln∣u+2(√6)+6∣ +ln(2(√6)−5)ln∣u−2(√6)+6∣+c =(1/(4(√6)))[Li_2 (((u+2(√6)+6)/(2(√6)+5)))−Li_2 (((u−2(√6)+6)/(2(√6)−5))) +ln(−2(√6)−5)ln∣u+2(√6)+6∣ +ln(2(√6)−5)ln∣u−2(√6)+6∣+c =(1/(4(√6)))[Li_2 (((e^(2ix) +2(√6)+5)/(2(√6)+5)))−Li_2 (((e^(2ix) −2(√6)+5)/(2(√6)−5))) +ln(−2(√6)−5)ln∣e^(2ix) +2(√6)+5∣ +ln(2(√6)−5)ln∣e^(2ix) −2(√6)+5∣+c](https://www.tinkutara.com/question/Q97452.png)
$${sinx}=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}} \\ $$$$\int\frac{{x}}{−\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{3}}{dx} \\ $$$$=−\mathrm{4}\int\frac{{x}}{\left({e}^{{ix}} −{e}^{−{ix}} \right)^{\mathrm{2}} +\mathrm{12}}{dx} \\ $$$$=−\mathrm{4}\int\frac{{xe}^{\mathrm{2}{ix}} }{\left({e}^{\mathrm{2}{ix}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}{e}^{\mathrm{2}{ix}} }{dx} \\ $$$${let}\:{u}={e}^{\mathrm{2}{ix}} −\mathrm{1},\:{dx}=\frac{{du}}{\mathrm{2}{ie}^{\mathrm{2}{ix}} } \\ $$$${x}=−\frac{{iln}\left({u}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$−\int\frac{{ln}\left({u}+\mathrm{1}\right)}{{u}^{\mathrm{2}} +\mathrm{12}+\mathrm{12}}{du} \\ $$$$=−\frac{{ln}\left({u}+\mathrm{1}\right)}{\left[{u}−\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{6}\right)\right]\left[{u}−\left(−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{6}\right]\right.}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{6}}}\left[\int\frac{{ln}\left({u}+\mathrm{1}\right)}{{u}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6}}{du}−\int\frac{{ln}\left({u}+\mathrm{1}\right)}{{u}−\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6}}{du}\right] \\ $$$${let}\:{v}={u}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6},{dv}={du} \\ $$$${let}\:{t}={u}−\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6},{dt}={du} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{6}}}\left[\int\frac{{ln}\left({v}−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right)}{{v}}{dv}−\int\frac{{ln}\left({t}+\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right)}{{t}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{6}}}\int\left[\frac{\:{ln}\left(\frac{{v}}{−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}}+\mathrm{1}\right)}{{v}}+{ln}\left(−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right)\frac{\mathrm{1}}{{v}}\right]{dv} \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{6}}}\int\left[\frac{{ln}\left(\frac{{t}}{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}}+\mathrm{1}\right)}{{t}}+{ln}\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right)\frac{\mathrm{1}}{{v}}\right]{dv} \\ $$$${let}\:{w}=−\frac{{v}}{−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}}\:,{dv}=\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right){dw} \\ $$$${let}\:{z}=−\frac{{t}}{−\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{5}}\:,{dt}=\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\right){dz} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{6}}}\left\{−\int−\frac{{ln}\left(\mathrm{1}−{w}\right)}{{w}}{dw}+\int−\frac{{ln}\left(\mathrm{1}−{z}\right)}{{z}}{dz}\right. \\ $$$$\left.+{ln}\left(−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right){ln}\mid{v}\mid+{ln}\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right){ln}\mid{t}\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{6}}}\left[{Li}_{\mathrm{2}} \left({w}\right)−{Li}_{\mathrm{2}} \left({z}\right)+{ln}\left(−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right){ln}\mid{u}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6}\mid\right. \\ $$$$+{ln}\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right){ln}\mid{u}−\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{6}}}\left[{Li}_{\mathrm{2}} \left(\frac{{u}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{5}}\right)−{Li}_{\mathrm{2}} \left(\frac{{u}−\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}}\right)\right. \\ $$$$+{ln}\left(−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right){ln}\mid{u}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6}\mid \\ $$$$+{ln}\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right){ln}\mid{u}−\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{6}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{6}}}\left[{Li}_{\mathrm{2}} \left(\frac{{e}^{\mathrm{2}{ix}} +\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{5}}\right)−{Li}_{\mathrm{2}} \left(\frac{{e}^{\mathrm{2}{ix}} −\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}}\right)\right. \\ $$$$+{ln}\left(−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right){ln}\mid{e}^{\mathrm{2}{ix}} +\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{5}\mid \\ $$$$+{ln}\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{5}\right){ln}\mid{e}^{\mathrm{2}{ix}} −\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{5}\mid+{c} \\ $$
Commented by student work last updated on 08/Jun/20
$$\mathrm{thanks}\:\mathrm{lot}\:\mathrm{very}\:\mathrm{good} \\ $$