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Question Number 40251 by MJS last updated on 17/Jul/18

1.     ∫(dα/(sin 2α +tan 3α))=?  2.     ∫(dβ/(cos 2β +cos 3β))=?  3.     ∫(dγ/(sinh 2γ +tanh 3γ))=?  4.     ∫(dδ/(cosh 2δ +cosh 3δ))=?

$$\mathrm{1}.\:\:\:\:\:\int\frac{{d}\alpha}{\mathrm{sin}\:\mathrm{2}\alpha\:+\mathrm{tan}\:\mathrm{3}\alpha}=? \\ $$$$\mathrm{2}.\:\:\:\:\:\int\frac{{d}\beta}{\mathrm{cos}\:\mathrm{2}\beta\:+\mathrm{cos}\:\mathrm{3}\beta}=? \\ $$$$\mathrm{3}.\:\:\:\:\:\int\frac{{d}\gamma}{\mathrm{sinh}\:\mathrm{2}\gamma\:+\mathrm{tanh}\:\mathrm{3}\gamma}=? \\ $$$$\mathrm{4}.\:\:\:\:\:\int\frac{{d}\delta}{\mathrm{cosh}\:\mathrm{2}\delta\:+\mathrm{cosh}\:\mathrm{3}\delta}=? \\ $$

Answered by maxmathsup by imad last updated on 26/Jul/18

4) let I = ∫   (dx/(ch(2x)+ch(3x)))  I = ∫    (dx/(((e^(2x)  +e^(−2x) )/2)+((e^(3x)  +e^(−3x) )/2))) =∫     ((2dx)/(e^(2x)  +e^(−2x)  +e^(3x)  +e^(−3x) ))  =_(e^x =t)     ∫   (2/(t^2  +t^(−2)  +t^3  +t^(−3) )) (dt/t) = ∫    ((2dt)/(t^3  +t^(−1)  +t^4  +t^(−2) ))  = ∫    ((t^2 dt)/(t^2 (t^3  +t^(−1)  +t^4  +t^(−2) ))) = ∫   ((t^2 dt)/(t^5  +t +t^6  +1)) =∫  ((t^2 dt)/(t^6  +t^5  +t+1))  roots of  p(t)=t^6  +t^5  +t+1  t_1 =−1 (double)  t_2 ∼0,809 +0,587i (complex)  t_3 =0,809−0,587i (complex)  t_4 ∼−0,309 +0,9511i(complex)  t_5  ∼ −0,309 −0,9511i(complex)  p(t) =(t+1)^2 ( t−t_2 )(t−t_2 ^− )(t−t_4 )(t−t_4 ^− ) ⇒  F(t)= (t^2 /((t+1)^2 (t^2 −2Re(t_2 )t  +∣t_2 ∣^2 )(t^2  −2Re(t_4 )t +∣t_4 ∣^2 )))  = (a/(t+1)) +(b/((t+1)^2 )) +((ct +d)/(t^2  −2Re(t_2 )t +∣t∣^2 )) + ((et +f)/(t^2 −2Re(t_4 )t +∣t_4 ∣^2 ))  ...be continued...

$$\left.\mathrm{4}\right)\:{let}\:{I}\:=\:\int\:\:\:\frac{{dx}}{{ch}\left(\mathrm{2}{x}\right)+{ch}\left(\mathrm{3}{x}\right)} \\ $$$${I}\:=\:\int\:\:\:\:\frac{{dx}}{\frac{{e}^{\mathrm{2}{x}} \:+{e}^{−\mathrm{2}{x}} }{\mathrm{2}}+\frac{{e}^{\mathrm{3}{x}} \:+{e}^{−\mathrm{3}{x}} }{\mathrm{2}}}\:=\int\:\:\:\:\:\frac{\mathrm{2}{dx}}{{e}^{\mathrm{2}{x}} \:+{e}^{−\mathrm{2}{x}} \:+{e}^{\mathrm{3}{x}} \:+{e}^{−\mathrm{3}{x}} } \\ $$$$=_{{e}^{{x}} ={t}} \:\:\:\:\int\:\:\:\frac{\mathrm{2}}{{t}^{\mathrm{2}} \:+{t}^{−\mathrm{2}} \:+{t}^{\mathrm{3}} \:+{t}^{−\mathrm{3}} }\:\frac{{dt}}{{t}}\:=\:\int\:\:\:\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{3}} \:+{t}^{−\mathrm{1}} \:+{t}^{\mathrm{4}} \:+{t}^{−\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\frac{{t}^{\mathrm{2}} {dt}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{3}} \:+{t}^{−\mathrm{1}} \:+{t}^{\mathrm{4}} \:+{t}^{−\mathrm{2}} \right)}\:=\:\int\:\:\:\frac{{t}^{\mathrm{2}} {dt}}{{t}^{\mathrm{5}} \:+{t}\:+{t}^{\mathrm{6}} \:+\mathrm{1}}\:=\int\:\:\frac{{t}^{\mathrm{2}} {dt}}{{t}^{\mathrm{6}} \:+{t}^{\mathrm{5}} \:+{t}+\mathrm{1}} \\ $$$${roots}\:{of}\:\:{p}\left({t}\right)={t}^{\mathrm{6}} \:+{t}^{\mathrm{5}} \:+{t}+\mathrm{1} \\ $$$${t}_{\mathrm{1}} =−\mathrm{1}\:\left({double}\right) \\ $$$${t}_{\mathrm{2}} \sim\mathrm{0},\mathrm{809}\:+\mathrm{0},\mathrm{587}{i}\:\left({complex}\right) \\ $$$${t}_{\mathrm{3}} =\mathrm{0},\mathrm{809}−\mathrm{0},\mathrm{587}{i}\:\left({complex}\right) \\ $$$${t}_{\mathrm{4}} \sim−\mathrm{0},\mathrm{309}\:+\mathrm{0},\mathrm{9511}{i}\left({complex}\right) \\ $$$${t}_{\mathrm{5}} \:\sim\:−\mathrm{0},\mathrm{309}\:−\mathrm{0},\mathrm{9511}{i}\left({complex}\right) \\ $$$${p}\left({t}\right)\:=\left({t}+\mathrm{1}\right)^{\mathrm{2}} \left(\:{t}−{t}_{\mathrm{2}} \right)\left({t}−\overset{−} {{t}}_{\mathrm{2}} \right)\left({t}−{t}_{\mathrm{4}} \right)\left({t}−\overset{−} {{t}}_{\mathrm{4}} \right)\:\Rightarrow \\ $$$${F}\left({t}\right)=\:\frac{{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{Re}\left({t}_{\mathrm{2}} \right){t}\:\:+\mid{t}_{\mathrm{2}} \mid^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({t}_{\mathrm{4}} \right){t}\:+\mid{t}_{\mathrm{4}} \mid^{\mathrm{2}} \right)} \\ $$$$=\:\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{b}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({t}_{\mathrm{2}} \right){t}\:+\mid{t}\mid^{\mathrm{2}} }\:+\:\frac{{et}\:+{f}}{{t}^{\mathrm{2}} −\mathrm{2}{Re}\left({t}_{\mathrm{4}} \right){t}\:+\mid{t}_{\mathrm{4}} \mid^{\mathrm{2}} }\:\:...{be}\:{continued}... \\ $$

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