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Question Number 58488 by Mr X pcx last updated on 23/Apr/19
let f(x) =∫ (dt/(x +cost +cos(2t))) (x real) 1) find a explicit form of f(x) 2)determine also ∫ (dt/((x+cost +cos(2t))^2 )) 3) find ∫ (dt/(1+cos(t)+cos(2t))) and ∫ (dt/((3 +cos(t)+cos(2t))^2 ))
$${let}\:{f}\left({x}\right)\:=\int\:\:\:\frac{{dt}}{{x}\:+{cost}\:+{cos}\left(\mathrm{2}{t}\right)}\:\:\left({x}\:{real}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){determine}\:{also}\:\int\:\:\frac{{dt}}{\left({x}+{cost}\:+{cos}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:\int\:\:\:\frac{{dt}}{\mathrm{1}+{cos}\left({t}\right)+{cos}\left(\mathrm{2}{t}\right)}\:{and} \\ $$$$\int\:\:\:\frac{{dt}}{\left(\mathrm{3}\:+{cos}\left({t}\right)+{cos}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 25/Apr/19
1) we have f(x) =∫ (dt/(x+cost +2cos^2 t−1)) let put cost =u and decompose F(u) =(1/(2u^2 +u +x−1)) ⇒Δ =1−8(x−1) =9−8x case 1 9−8x<0 ⇒no roots and (1/(2u^2 +u +x−1)) =(1/(2{u^2 +(u/2) +((x−1)/2)})) =(1/(2{ u^2 +2(1/4)u +(1/(16)) +((x−1)/2)−(1/(16))})) =(1/(2{ (u+(1/2))^2 +((8x−9)/(16))})) ⇒f(x) =∫ (dt/(2{(cost +(1/2))^2 +((8x−9)/(16))})) changement cost +(1/2) =((√(8x−9))/4)u give 4cost +2 =(√(8x−9))u ⇒ u =(1/( (√(8x−9))))(4cost +2) ⇒du =((−4 sint)/( (√(8x−9))))dt =((−4(√(1−cos^2 t)))/( (√(8x−9)))) dt ⇒ (dt/du) =−((√(8x−9))/(4(√(1−(((√(8x−9))/4)u−(1/2))^2 )))) ⇒ f(x) = ∫ ((√(8x−9))/(8(√(1−(((√(8x−9))/4)u−(1/2))^2 ))(((8x−9)/(16)))(1+u^2 ))) du ...be continued... case 2 if 9−8x>0 ⇒F(u) have two poles u_1 =((−1+(√(9−8x)))/4) u_2 =((−1−(√(9−8x)))/4) and F(u) =(1/(2(u−u_1 )(u−u_2 ))) =(1/(2(u_1 −u_2 ))){(1/(u−u_1 )) −(1/(u−u_2 ))} = (1/(2(((√(9−8x))/2)))){ (1/(u−u_1 )) −(1/(u−u_2 ))} ⇒ f(x) =(1/( (√(9−8x)))){ ∫ (dt/(cost −u_1 )) −∫ (dt/(cost −u_2 ))}let find ∫ (dt/(cost −a)) ch. tan((t/2)) =u give ∫ (dt/(cost −a)) =∫ (1/(((1−u^2 )/(1+u^2 )) −a)) ((2du)/(1+u^2 )) = ∫ ((2du)/(1−u^2 −a(1+u^2 ))) =∫ ((2du)/(1−u^2 −a−au^2 )) =∫ ((2du)/(1−a−(1+a)u^2 )) =∫ ((2du)/((1+a)u^2 +a−1)) ...
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\int\:\:\frac{{dt}}{{x}+{cost}\:+\mathrm{2}{cos}^{\mathrm{2}} {t}−\mathrm{1}}\:{let}\:{put}\:{cost}\:={u}\:{and}\:{decompose} \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} \:+{u}\:+{x}−\mathrm{1}}\:\Rightarrow\Delta\:=\mathrm{1}−\mathrm{8}\left({x}−\mathrm{1}\right)\:=\mathrm{9}−\mathrm{8}{x}\:\:\:{case}\:\mathrm{1}\:\:\mathrm{9}−\mathrm{8}{x}<\mathrm{0}\:\Rightarrow{no}\:{roots} \\ $$$${and}\:\:\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} \:+{u}\:+{x}−\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}\left\{{u}^{\mathrm{2}} \:+\frac{{u}}{\mathrm{2}}\:+\frac{{x}−\mathrm{1}}{\mathrm{2}}\right\}}\:=\frac{\mathrm{1}}{\mathrm{2}\left\{\:{u}^{\mathrm{2}} \:+\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}}{u}\:\:+\frac{\mathrm{1}}{\mathrm{16}}\:+\frac{{x}−\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{16}}\right\}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left\{\:\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{8}{x}−\mathrm{9}}{\mathrm{16}}\right\}}\:\Rightarrow{f}\left({x}\right)\:=\int\:\:\:\:\frac{{dt}}{\mathrm{2}\left\{\left({cost}\:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{8}{x}−\mathrm{9}}{\mathrm{16}}\right\}} \\ $$$${changement}\:\:{cost}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{8}{x}−\mathrm{9}}}{\mathrm{4}}{u}\:{give}\:\:\mathrm{4}{cost}\:+\mathrm{2}\:=\sqrt{\mathrm{8}{x}−\mathrm{9}}{u}\:\Rightarrow \\ $$$${u}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}{x}−\mathrm{9}}}\left(\mathrm{4}{cost}\:+\mathrm{2}\right)\:\:\Rightarrow{du}\:=\frac{−\mathrm{4}\:{sint}}{\:\sqrt{\mathrm{8}{x}−\mathrm{9}}}{dt}\:=\frac{−\mathrm{4}\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {t}}}{\:\sqrt{\mathrm{8}{x}−\mathrm{9}}}\:{dt}\:\Rightarrow \\ $$$$\frac{{dt}}{{du}}\:=−\frac{\sqrt{\mathrm{8}{x}−\mathrm{9}}}{\mathrm{4}\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{8}{x}−\mathrm{9}}}{\mathrm{4}}{u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\int\:\:\:\:\frac{\sqrt{\mathrm{8}{x}−\mathrm{9}}}{\mathrm{8}\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{8}{x}−\mathrm{9}}}{\mathrm{4}}{u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\left(\frac{\mathrm{8}{x}−\mathrm{9}}{\mathrm{16}}\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:{du}\:\:\:…{be}\:{continued}… \\ $$$${case}\:\mathrm{2}\:\:\:{if}\:\mathrm{9}−\mathrm{8}{x}>\mathrm{0}\:\Rightarrow{F}\left({u}\right)\:{have}\:{two}\:{poles}\:\:{u}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{9}−\mathrm{8}{x}}}{\mathrm{4}} \\ $$$${u}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{9}−\mathrm{8}{x}}}{\mathrm{4}}\:\:{and}\:{F}\left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)}\left\{\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right\}\:=\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\sqrt{\mathrm{9}−\mathrm{8}{x}}}{\mathrm{2}}\right)}\left\{\:\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{9}−\mathrm{8}{x}}}\left\{\:\:\int\:\:\:\:\frac{{dt}}{{cost}\:−{u}_{\mathrm{1}} }\:−\int\:\frac{{dt}}{{cost}\:−{u}_{\mathrm{2}} }\right\}{let}\:{find}\:\int\:\:\:\frac{{dt}}{{cost}\:−{a}} \\ $$$${ch}.\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}\:\:{give}\:\int\:\:\:\frac{{dt}}{{cost}\:−{a}}\:=\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:−{a}}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} \:−{a}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\int\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} −{a}−{au}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{du}}{\mathrm{1}−{a}−\left(\mathrm{1}+{a}\right){u}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{a}\right){u}^{\mathrm{2}} +{a}−\mathrm{1}}\:\:… \\ $$
Commented by maxmathsup by imad last updated on 26/Apr/19
⇒I=∫ ((2du)/((1+a)u^2 +a−1)) =(2/(1+a)) ∫ (du/(u^2 +((a−1)/(a+1)))) if ((a−1)/(a+1))>0 we do the changement u =(√((a−1)/(a+1)))t ⇒ I = (2/(1+a)) ((a+1)/(a−1)) ∫ (1/(t^2 +1)) (√((a−1)/(a+1)))dt =(2/( (√(a^2 −1)))) arctan((√((a+1)/(a−1)))u) +c_1 =(2/( (√(a^2 −1)))) arctan( (√((a+1)/(a−1)))tan((t/2)))+c_1 ....
$$\:\:\:\Rightarrow{I}=\int\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{a}\right){u}^{\mathrm{2}} \:+{a}−\mathrm{1}}\:=\frac{\mathrm{2}}{\mathrm{1}+{a}}\:\int\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}\:\:{if}\:\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}>\mathrm{0}\:\:{we}\:{do}\:{the}\:{changement} \\ $$$${u}\:=\sqrt{\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}{t}\:\Rightarrow\:\:{I}\:=\:\frac{\mathrm{2}}{\mathrm{1}+{a}}\:\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}\:\int\:\:\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\sqrt{\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}{dt} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:{arctan}\left(\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}{u}\right)\:+{c}_{\mathrm{1}} =\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:{arctan}\left(\:\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}{tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)+{c}_{\mathrm{1}} …. \\ $$

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