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Question Number 143032 by ZiYangLee last updated on 09/Jun/21

∫ sin^(−5) x dx =?

$$\:\:\:\:\:\int\:\mathrm{sin}^{−\mathrm{5}} {x}\:{dx}\:=? \\ $$

Answered by Olaf_Thorendsen last updated on 09/Jun/21

F(x) = ∫(dx/(sin^5 x)) Let t = tan(x/2) F(t) = ∫(1/((((2t)/(1+t^2 )))^5 )).((2dt)/(1+t^2 )) F(t) = (1/(16))∫(((1+t^2 )^4 )/t^5 ) dt F(t) = (1/(16))∫((1+4t^2 +6t^4 +4t^6 +t^8 )/t^5 ) dt F(t) = (1/(16))∫((1/t^5 )+(4/t^3 )+(6/t)+4t+t^3 ) dt F(t) = (1/(16))(−(1/(4t^4 ))−(2/t^2 )+6lnt+2t^2 +(t^4 /4))+C F(x) = −(1/(64tan^4 (x/2)))−(1/(8tan^2 (x/2)))+ (3/8)ln(tan(x/2))+(1/8)tan^2 (x/2)+((tan^4 (x/2))/(64))+C F(x) = (1/8)ln(tan(x/2))−((1/(4sin^4 x))+(3/(8sin^2 x)))cosx+C

$$\mathrm{F}\left({x}\right)\:=\:\int\frac{{dx}}{\mathrm{sin}^{\mathrm{5}} {x}} \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\int\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{5}} }.\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{4}} }{{t}^{\mathrm{5}} }\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int\frac{\mathrm{1}+\mathrm{4}{t}^{\mathrm{2}} +\mathrm{6}{t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{6}} +{t}^{\mathrm{8}} }{{t}^{\mathrm{5}} }\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{5}} }+\frac{\mathrm{4}}{{t}^{\mathrm{3}} }+\frac{\mathrm{6}}{{t}}+\mathrm{4}{t}+{t}^{\mathrm{3}} \right)\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\left(−\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{4}} }−\frac{\mathrm{2}}{{t}^{\mathrm{2}} }+\mathrm{6ln}{t}+\mathrm{2}{t}^{\mathrm{2}} +\frac{{t}^{\mathrm{4}} }{\mathrm{4}}\right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{64tan}^{\mathrm{4}} \frac{{x}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{8tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}+ \\ $$$$\frac{\mathrm{3}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\frac{\mathrm{tan}^{\mathrm{4}} \frac{{x}}{\mathrm{2}}}{\mathrm{64}}+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\right)−\left(\frac{\mathrm{1}}{\mathrm{4sin}^{\mathrm{4}} {x}}+\frac{\mathrm{3}}{\mathrm{8sin}^{\mathrm{2}} {x}}\right)\mathrm{cos}{x}+\mathrm{C} \\ $$

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