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Question Number 36436 by prof Abdo imad last updated on 02/Jun/18

find ∫ ((sinx)/(1+cos^3 x))dx

$${find}\:\:\int\:\:\:\:\:\frac{{sinx}}{\mathrm{1}+{cos}^{\mathrm{3}} {x}}{dx} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

=∫((−dt)/((1+t)(1−t+t^2 ))) t=cosx =(1/((1+t)(1−t+t^2 )))=(a/(1+t))+((bt+c)/(1−t+t^2 )) 1=a−at+at^2 +bt+c+bt^2 +ct 1=a+c+t(−a+b+c)+t^2 (a+b) a+c=1 −a+b+c=0 a+b=0 −a−a+1−a=0 −3a=−1 a=(1/3) b=−(1/3) c^ =1−(1/3) a=(1/3) b=−(1/3) c=(2/3) =−1∫((1/3)/(1+t))dt−∫((((−1)/3)t+(2/3))/(1−t+t^2 ))dt =((−1)/3)ln(1+t)+(1/3)∫((t−2)/(1−t+t^2 ))dt =do+(1/6)∫((2t−4)/(t^2 −t+1)) =do+(1/6)∫((2t−1)/(t^2 −t+1))dt−(1/6)∫(3/(t^2 −t+1))dt =do+(1/6)ln(t^2 −t+1)−(3/6)∫(dt/(t^2 −2t.(1/2)+(1/4)+1−(1/4))) =((−1)/3)ln(1+t)+(1/6)ln(t^2 −t+1)−(3/6)∫(dt/((t−(1/2))^2 +(3/4))) do−(1/2)×(2/(√3))tan^(−1) (((t−(1/2))/((√3)/2)))

$$=\int\frac{−{dt}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)}\:\:\:\:{t}={cosx} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)}=\frac{{a}}{\mathrm{1}+{t}}+\frac{{bt}+{c}}{\mathrm{1}−{t}+{t}^{\mathrm{2}} } \\ $$$$\:\mathrm{1}={a}−{at}+{at}^{\mathrm{2}} +{bt}+{c}+{bt}^{\mathrm{2}} +{ct} \\ $$$$ \\ $$$$\mathrm{1}={a}+{c}+{t}\left(−{a}+{b}+{c}\right)+{t}^{\mathrm{2}} \left({a}+{b}\right) \\ $$$${a}+{c}=\mathrm{1} \\ $$$$−{a}+{b}+{c}=\mathrm{0} \\ $$$${a}+{b}=\mathrm{0} \\ $$$$−{a}−{a}+\mathrm{1}−{a}=\mathrm{0} \\ $$$$−\mathrm{3}{a}=−\mathrm{1} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\overset{} {{c}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${c}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$=−\mathrm{1}\int\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}+{t}}{dt}−\int\frac{\frac{−\mathrm{1}}{\mathrm{3}}{t}+\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}−{t}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{1}+{t}\right)+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{t}−\mathrm{2}}{\mathrm{1}−{t}+{t}^{\mathrm{2}} }{dt} \\ $$$$={do}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{t}−\mathrm{4}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}} \\ $$$$={do}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{3}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$={do}+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{6}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}{t}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{1}+{t}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{6}}\int\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$${do}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right) \\ $$$$ \\ $$

Answered by MJS last updated on 02/Jun/18

∫((sin x)/(1+cos^3 x))dx= [((t=2arctan x → dx=((2dt)/(1+t^2 )))),((sin x=((2t)/(1+t^2 )) cos x=((1−t^2 )/(1+t^2 )))) ] =2∫((t^3 +t)/(3t^4 +1))dt= [u=t^2 → dt=(du/(2t))] =∫((u+1)/(3u^2 +1))du=∫(u/(3u^2 +1))du+∫(du/(3u^2 +1))= [v=3u^2 +1 → du=(6/dv); w=(√3)u → du=((√3)/3)dw] =(1/6)∫(dv/v)+((√3)/3)∫(dw/(w^2 +1))= =(1/6)ln v+((√3)/3)arctan w= =(1/6)ln(3u^2 +1)+((√3)/3)arctan((√3)u)= =(1/6)ln(3t^4 +1)+((√3)/3)arctan((√3)t^2 )= =(1/6)ln(48(1+arctan^4 x))+((√3)/3)arctan(4(√3)arctan^2 x)+C

$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{3}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{{t}=\mathrm{2arctan}\:{x}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}\\{\mathrm{sin}\:{x}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\mathrm{cos}\:{x}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\end{bmatrix} \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{3}} +{t}}{\mathrm{3}{t}^{\mathrm{4}} +\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}={t}^{\mathrm{2}} \:\rightarrow\:{dt}=\frac{{du}}{\mathrm{2}{t}}\right] \\ $$$$=\int\frac{{u}+\mathrm{1}}{\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}}{du}=\int\frac{{u}}{\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}}{du}+\int\frac{{du}}{\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{v}=\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}\:\rightarrow\:{du}=\frac{\mathrm{6}}{{dv}};\:{w}=\sqrt{\mathrm{3}}{u}\:\rightarrow\:{du}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{dw}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{dv}}{{v}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\int\frac{{dw}}{{w}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:{v}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:{w}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left(\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\left(\sqrt{\mathrm{3}}{u}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left(\mathrm{3}{t}^{\mathrm{4}} +\mathrm{1}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\left(\sqrt{\mathrm{3}}{t}^{\mathrm{2}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left(\mathrm{48}\left(\mathrm{1}+\mathrm{arctan}^{\mathrm{4}} \:{x}\right)\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\left(\mathrm{4}\sqrt{\mathrm{3}}\mathrm{arctan}^{\mathrm{2}} \:{x}\right)+{C} \\ $$

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