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Question Number 47059 by maxmathsup by imad last updated on 04/Nov/18
find ∫ (dx/(1+cos x +cos(2x)))
$$\:{find}\:\int\:\:\frac{{dx}}{\mathrm{1}+{cos}\:{x}\:+{cos}\left(\mathrm{2}{x}\right)} \\ $$
Commented by maxmathsup by imad last updated on 06/Nov/18
let A =∫ (dx/(1+cosx +cos(2x))) = ∫ (dx/(1+cos(x)+2cos^2 x−1)) =∫ (dx/(cosx(1+2cosx))) =∫ ((1/(cosx)) −(2/(1+2cosx)))dx =∫ (dx/(cosx)) −∫ (2/(1+2cosx))dx but ∫ (dx/(cosx)) =_(tan((x/2))=t) ∫ (1/((1−t^2 )/(1+t^2 ))) ((2dt)/(1+t^2 )) =∫ ((2dt)/(1−t^2 )) =∫ ((1/(1−t)) +(1/(1+t)))dt =ln∣((1+t)/(1−t))∣ +c_1 =ln∣((1+tan((x/2)))/(1−tan((x/2))))∣=ln∣tan((x/2)+(π/4))∣ +c_1 also we have ∫ (2/(1+2cosx))dt =_(tan((x/2))=t) ∫ (2/(1+2 ((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =4 ∫ (dt/(1+t^2 +2−2t^2 )) =4 ∫ (dt/(3−t^2 )) =_(t =(√3)u) 4 ∫ (((√3)du)/(3(1−u^2 ))) =((2(√3))/4) ∫ ((2du)/(1−u^2 )) =((√3)/2)ln∣((1+u)/(1−u))∣ +c_2 =((√3)/2)ln∣((1+(t/( (√3))))/(1−(t/( (√3)))))∣ +c_2 =((√3)/2)ln∣(((√3)+t)/( (√3)−t))∣ +c_2 finally A =ln∣tan((x/2)+(π/4))∣−((√3)/2)ln∣(((√3)+tan((x/2)))/( (√3)−tan((x/2))))∣ +C .
$${let}\:{A}\:=\int\:\:\:\frac{{dx}}{\mathrm{1}+{cosx}\:+{cos}\left(\mathrm{2}{x}\right)}\:=\:\int\:\:\frac{{dx}}{\mathrm{1}+{cos}\left({x}\right)+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}} \\ $$$$=\int\:\:\frac{{dx}}{{cosx}\left(\mathrm{1}+\mathrm{2}{cosx}\right)}\:=\int\:\left(\frac{\mathrm{1}}{{cosx}}\:−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}{cosx}}\right){dx}\:=\int\:\frac{{dx}}{{cosx}}\:−\int\:\:\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}{cosx}}{dx} \\ $$$${but}\:\int\:\:\frac{{dx}}{{cosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\int\:\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\ $$$$={ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:+{c}_{\mathrm{1}} ={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid\:+{c}_{\mathrm{1}} \:\:{also}\:{we}\:{have} \\ $$$$\int\:\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}{cosx}}{dt}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{4}\:\int\:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int\:\:\frac{{dt}}{\mathrm{3}−{t}^{\mathrm{2}} }\:=_{{t}\:=\sqrt{\mathrm{3}}{u}} \:\mathrm{4}\:\:\int\:\:\:\:\:\frac{\sqrt{\mathrm{3}}{du}}{\mathrm{3}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\int\:\:\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\:+{c}_{\mathrm{2}} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+\frac{{t}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}−\frac{{t}}{\:\sqrt{\mathrm{3}}}}\mid\:+{c}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ln}\mid\frac{\sqrt{\mathrm{3}}+{t}}{\:\sqrt{\mathrm{3}}−{t}}\mid\:+{c}_{\mathrm{2}} \:\:{finally}\: \\ $$$${A}\:={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ln}\mid\frac{\sqrt{\mathrm{3}}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{3}}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid\:+{C}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18
∫(dx/(cosx+2cos^2 x)) ∫(dx/(cosx(1+2cosx))) ∫((1+2cosx−2cosx)/(cosx(1+2cosx)))dx ∫(dx/(cosx))−∫((2dx)/(2((1/2)+cosx))) ∫secxdx−∫(dx/((1/2)+((1−tan^2 (x/2))/(1+tan^2 (x/2))))) ∫secx dx−∫((sec^2 (x/2))/((3/2)−(1/2)tan^2 (x/2)))dx ∫secxdx−2∫((sec^2 (x/2))/(3−tan^2 (x/2)))dx I_1 −2I_2 I_1 =∫secxdx =ln∣sex+tanx∣+c_1 I_2 =∫((sec^2 (x/2))/(3−tan^2 (x/2)))dx t=tan(x/2) dt=sec^2 (x/2)×(1/2)dx I_2 =∫((2dt)/(((√3) )^2 −t^2 )) =2×(1/(2(√3)))ln∣(((√3) +t)/( (√3) −t))∣+c_2 =(1/( (√3)))ln∣(((√3) +tan(x/2))/( (√3) −tan(x/2)))∣+c_2
$$\int\frac{{dx}}{{cosx}+\mathrm{2}{cos}^{\mathrm{2}} {x}} \\ $$$$\int\frac{{dx}}{{cosx}\left(\mathrm{1}+\mathrm{2}{cosx}\right)} \\ $$$$\int\frac{\mathrm{1}+\mathrm{2}{cosx}−\mathrm{2}{cosx}}{{cosx}\left(\mathrm{1}+\mathrm{2}{cosx}\right)}{dx} \\ $$$$\int\frac{{dx}}{{cosx}}−\int\frac{\mathrm{2}{dx}}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+{cosx}\right)} \\ $$$$\int{secxdx}−\int\frac{{dx}}{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\int{secx}\:{dx}−\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\int{secxdx}−\mathrm{2}\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{3}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$${I}_{\mathrm{1}} −\mathrm{2}{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int{secxdx} \\ $$$$={ln}\mid{sex}+{tanx}\mid+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{3}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$${t}={tan}\frac{{x}}{\mathrm{2}}\:\:\:{dt}={sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}{dx} \\ $$$${I}_{\mathrm{2}} =\int\frac{\mathrm{2}{dt}}{\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} −{t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\mid\frac{\sqrt{\mathrm{3}}\:+{t}}{\:\sqrt{\mathrm{3}}\:−{t}}\mid+{c}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{ln}\mid\frac{\sqrt{\mathrm{3}}\:+{tan}\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{3}}\:−{tan}\frac{{x}}{\mathrm{2}}}\mid+{c}_{\mathrm{2}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 05/Nov/18
thanks sir.
$${thanks}\:{sir}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18
thank you sir...
$${thank}\:{you}\:{sir}… \\ $$
Answered by ajfour last updated on 04/Nov/18
I=∫(dx/(2cos^2 x+cos x)) let tan (x/2) = t ⇒ dx = ((2dt)/(1+t^2 )) I=∫((2dt)/((1+t^2 )[2(((1−t^2 )/(1+t^2 )))^2 +((1−t^2 )/(1+t^2 ))])) =∫((2(1+t^2 )dt)/(2(1−t^2 )^2 +1−t^4 )) = 2∫(((1+t^2 )dt)/(t^4 −4t^2 +3)) = 2∫ ((((1/t^2 )+1)dt)/(t^2 +(3/t^2 )−4)) let a(1−((√3)/t^2 ))+b(1+((√3)/t^2 ))=(1/t^2 )+1 ⇒ a+b=1 & b−a = (1/( (√3))) ⇒ a=(1/2)−(1/(2(√3))) ; b = (1/2)+(1/(2(√3))) I= (1−(1/( (√3))))∫(((1−((√3)/t^2 ))dt)/((t+((√3)/t))^2 −4−2(√3))) +(1+(1/( (√3))))∫(((1+((√3)/t^2 ))dt)/((t−((√3)/t))^2 −4+2(√3))) I= ((((√3)−1)/( (√3))))(1/(2(√(4+2(√3)))))ln ∣((t+((√3)/t)−(√(4+2(√3))))/(t+((√3)/t)+(√(4+2(√3)))))∣ +((((√3)+1)/( (√3))))(1/(2(√(4−2(√3)))))ln ∣((t+((√3)/t)−(√(4−2(√3))))/(t+((√3)/t)+(√(4−2(√3)))))∣+c .
$${I}=\int\frac{{dx}}{\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:{x}} \\ $$$$\:\:\:{let}\:\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\:{t}\:\:\Rightarrow\:\:{dx}\:=\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${I}=\int\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left[\mathrm{2}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right]} \\ $$$$\:\:\:=\int\frac{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt}}{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{1}−{t}^{\mathrm{4}} } \\ $$$$\:\:=\:\mathrm{2}\int\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt}}{{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}}\:=\:\mathrm{2}\int\:\frac{\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}\right){dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{3}}{{t}^{\mathrm{2}} }−\mathrm{4}} \\ $$$${let}\:{a}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{{t}^{\mathrm{2}} }\right)+{b}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{{t}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1} \\ $$$$\Rightarrow\:\:{a}+{b}=\mathrm{1}\:\:\:\&\:\:{b}−{a}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\:{a}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:;\:\:{b}\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\:\:{I}=\:\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\int\frac{\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{{t}^{\mathrm{2}} }\right){dt}}{\left({t}+\frac{\sqrt{\mathrm{3}}}{{t}}\right)^{\mathrm{2}} −\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:+\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\int\frac{\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{{t}^{\mathrm{2}} }\right){dt}}{\left({t}−\frac{\sqrt{\mathrm{3}}}{{t}}\right)^{\mathrm{2}} −\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$${I}=\:\left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}}\mathrm{ln}\:\mid\frac{{t}+\frac{\sqrt{\mathrm{3}}}{{t}}−\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}}{{t}+\frac{\sqrt{\mathrm{3}}}{{t}}+\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}}\mid \\ $$$$\:+\left(\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}}\mathrm{ln}\:\mid\frac{{t}+\frac{\sqrt{\mathrm{3}}}{{t}}−\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}}{{t}+\frac{\sqrt{\mathrm{3}}}{{t}}+\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}}\mid+{c}\:. \\ $$
Commented by maxmathsup by imad last updated on 04/Nov/18
thank you sir Ajfour
$${thank}\:{you}\:{sir}\:{Ajfour} \\ $$

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