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Question Number 202125 by Calculusboy last updated on 21/Dec/23
∫ ((sin(3x))/(1+sin^3 x))dx
$$\int\:\frac{\boldsymbol{{sin}}\left(\mathrm{3}\boldsymbol{{x}}\right)}{\mathrm{1}+\boldsymbol{{sin}}^{\mathrm{3}} \boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$
Answered by Frix last updated on 21/Dec/23
Let s=sin x ((sin 3x)/(1+sin^3 x))=((s(4s^2 −3))/((s+1)(s^2 −s+1)))= =−4+(1/(3(s+1)))−((s−11)/(3(s^2 −s+1))) ∫((sin 3x)/(1+sin^3 x))dx= =−4∫dx+(1/3)∫(dx/(1+sin x))+(1/3)∫((11−sin x)/(1−sin x +sin^2 x))dx These are easy: −4∫dx=−4x (1/3)∫(dx/(1+sin x))=−((1−sin x)/(3cos x)) This is unpleasant: (1/3)∫((11−sin x)/(1−sin x +sin^2 x))dx I tried t=tan (x/2) but... Instead I solved it with t=tan x +(1/(cos x)) ⇔ x=tan^(−1) ((t^2 −1)/(2t)) → dx=((2dt)/(t^2 +1)) which leads to (4/3)∫((5t^2 +6)/(t^4 +3))dt=(4/3)∫((5t^2 +6)/((t^2 −((12))^(1/4) t+(√3))(t^2 +((12))^(1/4) t+(√3))))dt Now decompose...
$$\mathrm{Let}\:{s}=\mathrm{sin}\:{x} \\ $$$$\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{3}} \:{x}}=\frac{{s}\left(\mathrm{4}{s}^{\mathrm{2}} −\mathrm{3}\right)}{\left({s}+\mathrm{1}\right)\left({s}^{\mathrm{2}} −{s}+\mathrm{1}\right)}= \\ $$$$=−\mathrm{4}+\frac{\mathrm{1}}{\mathrm{3}\left({s}+\mathrm{1}\right)}−\frac{{s}−\mathrm{11}}{\mathrm{3}\left({s}^{\mathrm{2}} −{s}+\mathrm{1}\right)} \\ $$$$\int\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{3}} \:{x}}{dx}= \\ $$$$=−\mathrm{4}\int{dx}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{11}−\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\mathrm{These}\:\mathrm{are}\:\mathrm{easy}: \\ $$$$−\mathrm{4}\int{dx}=−\mathrm{4}{x} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}}=−\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{3cos}\:{x}} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{unpleasant}: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{11}−\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\mathrm{I}\:\mathrm{tried}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{but}… \\ $$$$\mathrm{Instead}\:\mathrm{I}\:\mathrm{solved}\:\mathrm{it}\:\mathrm{with} \\ $$$${t}=\mathrm{tan}\:{x}\:+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:\Leftrightarrow\:{x}=\mathrm{tan}^{−\mathrm{1}} \:\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{t}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{6}}{{t}^{\mathrm{4}} +\mathrm{3}}{dt}=\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{6}}{\left({t}^{\mathrm{2}} −\sqrt[{\mathrm{4}}]{\mathrm{12}}{t}+\sqrt{\mathrm{3}}\right)\left({t}^{\mathrm{2}} +\sqrt[{\mathrm{4}}]{\mathrm{12}}{t}+\sqrt{\mathrm{3}}\right)}{dt} \\ $$$$\mathrm{Now}\:\mathrm{decompose}… \\ $$
Commented by Calculusboy last updated on 22/Dec/23
nice solution
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}} \\ $$

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