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Question Number 84577 by msup trace by abdo last updated on 14/Mar/20
calculate ∫ (dx/(cosx +cos(2x)+cos(3x)))
$${calculate}\:\int\:\:\:\:\frac{{dx}}{{cosx}\:+{cos}\left(\mathrm{2}{x}\right)+{cos}\left(\mathrm{3}{x}\right)} \\ $$
Commented by jagoll last updated on 14/Mar/20
cos x+cos 3x+cos 2x = 2cos 2x cos x + cos 2x = cos 2x ( 2cos x +1) ∫ (( dx)/(cos 2x (2cos x +1) )) = let tan ((x/2)) = t
$$\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{3}{x}+\mathrm{cos}\:\mathrm{2}{x}\:=\: \\ $$$$\mathrm{2cos}\:\mathrm{2}{x}\:\mathrm{cos}\:{x}\:+\:\mathrm{cos}\:\mathrm{2}{x}\:=\: \\ $$$$\mathrm{cos}\:\mathrm{2}{x}\:\left(\:\mathrm{2cos}\:{x}\:+\mathrm{1}\right)\: \\ $$$$\int\:\frac{\:\:{dx}}{\mathrm{cos}\:\mathrm{2}{x}\:\left(\mathrm{2cos}\:{x}\:+\mathrm{1}\right)\:}\:=\: \\ $$$$\mathrm{let}\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)\:=\:{t}\: \\ $$
Commented by abdomathmax last updated on 14/Mar/20
we have cosx +cos(2x)+cos(3x) =cosx +cos(3x)+cos(2x)=2cos(2x)cosx +cosx =cosx(2cos(2x)+1) =cosx(2(2cos^2 x−1)−1) =cosx(4cos^2 x−3) ⇒ I =∫ (dx/(cosx(4cos^2 x−1))) let decompose f(t)=(1/(t(4t^2 −1)))=(1/(t(2t−1)(2t+1))) ⇒f(t)=(a/t) +(b/(2t−1)) +(c/(2t+1)) a=−1 b= (2/2)=1 c=((−2)/(−2))=1 ⇒f(t)=−(1/t) +(1/(2t−1)) +(1/(2t+1)) ⇒ I =−∫ (dx/(cosx)) +∫ (dx/(2cosx−1)) +∫ (dx/(2cosx +1)) ∫ (dx/(cosx)) =_(tan((x/2))=u) ∫ ((2du)/((1+u^2 )×((1−u^2 )/(1+u^2 )))) =∫ ((2du)/((1−u)(1+u))) =∫ ((1/(1−u))+(1/(1+u)))du =ln∣((1+u)/(1−u))∣ +c_0 =ln∣((1+tan((x/2)))/(1−tan((x/2))))∣=ln∣tan((x/2)+(π/4))∣ +c_0 ∫ (dx/(2cosx−1)) =_(tan((x/2))=u) ∫ ((2du)/((1+u^2 )(2((1−u^2 )/(1+u^2 ))−1))) =∫ ((2du)/(2−2u^2 −1−u^2 )) =∫ ((2du)/(−3u^2 +1)) =−2 ∫ (du/(3u^2 −1)) =−2 ∫ (du/(((√3)u+1)((√3)u−1))) = ∫ ((1/( (√3)u+1))−(1/( (√3)u−1)))du =(1/( (√3)))ln∣(((√3)u+1)/( (√3)u −1))∣ +c_1 =(1/( (√3)))ln∣(((√3)tan((x/2))+1)/( (√3)tan((x/2))−1))∣ +c_1 we folow the same manner to dtermine ∫ (dx/(2cosx +1))
$${we}\:{have}\:{cosx}\:+{cos}\left(\mathrm{2}{x}\right)+{cos}\left(\mathrm{3}{x}\right) \\ $$$$={cosx}\:+{cos}\left(\mathrm{3}{x}\right)+{cos}\left(\mathrm{2}{x}\right)=\mathrm{2}{cos}\left(\mathrm{2}{x}\right){cosx}\:+{cosx} \\ $$$$={cosx}\left(\mathrm{2}{cos}\left(\mathrm{2}{x}\right)+\mathrm{1}\right)\:={cosx}\left(\mathrm{2}\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)−\mathrm{1}\right) \\ $$$$={cosx}\left(\mathrm{4}{cos}^{\mathrm{2}} {x}−\mathrm{3}\right)\:\Rightarrow \\ $$$${I}\:=\int\:\:\:\frac{{dx}}{{cosx}\left(\mathrm{4}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)}\:{let}\:{decompose} \\ $$$${f}\left({t}\right)=\frac{\mathrm{1}}{{t}\left(\mathrm{4}{t}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{\mathrm{1}}{{t}\left(\mathrm{2}{t}−\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)}\:\Rightarrow{f}\left({t}\right)=\frac{{a}}{{t}}\:+\frac{{b}}{\mathrm{2}{t}−\mathrm{1}}\:+\frac{{c}}{\mathrm{2}{t}+\mathrm{1}} \\ $$$${a}=−\mathrm{1} \\ $$$${b}=\:\frac{\mathrm{2}}{\mathrm{2}}=\mathrm{1} \\ $$$${c}=\frac{−\mathrm{2}}{−\mathrm{2}}=\mathrm{1}\:\Rightarrow{f}\left({t}\right)=−\frac{\mathrm{1}}{{t}}\:+\frac{\mathrm{1}}{\mathrm{2}{t}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{t}+\mathrm{1}}\:\Rightarrow \\ $$$${I}\:=−\int\:\:\frac{{dx}}{{cosx}}\:+\int\:\:\frac{{dx}}{\mathrm{2}{cosx}−\mathrm{1}}\:+\int\:\:\frac{{dx}}{\mathrm{2}{cosx}\:+\mathrm{1}} \\ $$$$\int\:\:\frac{{dx}}{{cosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}} \:\:\:\int\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)×\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$$=\int\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{u}\right)}\:=\int\:\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}+\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du} \\ $$$$={ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\:+\underset{\mathrm{0}} {{c}}={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid\:+{c}_{\mathrm{0}} \\ $$$$\int\:\:\frac{{dx}}{\mathrm{2}{cosx}−\mathrm{1}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}} \:\:\int\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }−\mathrm{1}\right)} \\ $$$$=\int\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}−\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}−{u}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{du}}{−\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=−\mathrm{2}\:\int\:\:\:\frac{{du}}{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}}\:=−\mathrm{2}\:\int\:\:\frac{{du}}{\left(\sqrt{\mathrm{3}}{u}+\mathrm{1}\right)\left(\sqrt{\mathrm{3}}{u}−\mathrm{1}\right)} \\ $$$$=\:\int\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}{u}+\mathrm{1}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}{u}−\mathrm{1}}\right){du} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{ln}\mid\frac{\sqrt{\mathrm{3}}{u}+\mathrm{1}}{\:\sqrt{\mathrm{3}}{u}\:−\mathrm{1}}\mid\:+{c}_{\mathrm{1}} \:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{ln}\mid\frac{\sqrt{\mathrm{3}}{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}}\mid\:\:+{c}_{\mathrm{1}} \\ $$$${we}\:{folow}\:{the}\:{same}\:{manner}\:{to}\:{dtermine} \\ $$$$\int\:\:\frac{{dx}}{\mathrm{2}{cosx}\:+\mathrm{1}} \\ $$

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