Question Number 51824 by Abdo msup. last updated on 30/Dec/18
$${find}\:\int\:\:\:\frac{{dx}}{{cosx}\:+{cos}\left(\mathrm{2}{x}\right)+{cos}\left(\mathrm{3}{x}\right)} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18
![∫(dx/(cos2x+2cos2x.cosx)) ∫(dx/(cos2x(1+2cosx))) ∫(dx/((2cos^2 x−1)(2cosx+1))) cosx=((1−t^2 )/(1+t^2 )) t=tan(x/2) (dt/dx)=(1/2)sec^2 (x/2) so ((2dt)/(1+t^2 ))=dx ∫(((2dt)/(1+t^2 ))/([2(((1−t^2 )/(1+t^2 )))^2 −1][2(((1−t^2 )/(1+t^2 )))+1])) ∫(((2dt)/(1+t^2 ))/([((2−4t^2 +2t^4 −1−2t^2 −t^4 )/((1+t^2 )^2 ))][((2−2t^2 +1+t^2 )/(1+t^2 ))])) ∫((((2(1+t^2 )^3 )/(1+t^2 ))dt)/((t^4 −6t^2 +1)(3−t^2 ))) 2∫(((1+t^2 )^2 )/((t^4 −6t^2 +1)(3−t^2 )))dt 2∫((t^4 +2t^2 +1)/((t^4 −6t^2 +1)(3−t^2 )))dt now t^4 +2t^2 +1=a(t^4 −6t^2 +1)+(bt^2 +c)(3−t^2 ) t^4 +2t^2 +1=at^4 −6at^2 +a+3bt^2 −bt^4 +3c−ct^2 t^4 +2t^2 +1=t^4 (a−b)+t^2 (−6a+3b−c)+(a+c) a−b=1 −6a+3b−c=2 a+c=1 −6a+3(a−1)−(1−a)=2 −6a+3a+a−3−1=2 −2a=6 a=−3 b=a−1=−4 c=1−a=4 a=−3 b=−4 c=4 2∫((a(t^4 −6t^2 +1)+(bt^2 +c)(3−t^2 ))/((t^4 −6t^2 +1)(3−t^2 )))dt 2∫((−3)/(3−t^2 ))dt+2∫(((−4t^2 +4))/(t^4 −6t^2 +1))dt −6∫(dt/(3−t^2 ))−8∫((t^2 −1)/(t^4 −6t^2 +1))dt 6∫(dt/(t^2 −3))−8∫((1−(1/t^2 ))/((t^2 +(1/t^2 ))−6)) 6∫(dt/(t^2 −3))−8∫((d(t+(1/t)))/((t+(1/t))^2 −8)) 6×(1/(2(√3)))ln(((t−(√3))/(t+(√3))))−8×(1/(2(√8) ))ln(((t+(1/t)−(√8))/(t+(1/t)+(√8))))+c =(√3) ln(((tan(x/2)−(√3))/(tan(x/2)+(√3))))−(√2) ln(((tan(x/2)+(1/(tan(x/2)))−2(√2))/(tan(x/2)+(1/(tan(x/2)))+2(√2))))+c](Q51836.png)
$$\int\frac{{dx}}{{cos}\mathrm{2}{x}+\mathrm{2}{cos}\mathrm{2}{x}.{cosx}} \\ $$$$\int\frac{{dx}}{{cos}\mathrm{2}{x}\left(\mathrm{1}+\mathrm{2}{cosx}\right)} \\ $$$$\int\frac{{dx}}{\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)\left(\mathrm{2}{cosx}+\mathrm{1}\right)} \\ $$$${cosx}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\:\:{t}={tan}\frac{{x}}{\mathrm{2}}\:\:\:\:\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}} \\ $$$${so}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }={dx} \\ $$$$\int\frac{\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\left[\mathrm{2}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{1}\right]\left[\mathrm{2}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)+\mathrm{1}\right]} \\ $$$$\int\frac{\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\left[\frac{\mathrm{2}−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} −\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} −{t}^{\mathrm{4}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]\left[\frac{\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right]} \\ $$$$\int\frac{\frac{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}}{\left({t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}−{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{2}\int\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\mathrm{2}\int\frac{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{dt} \\ $$$${now} \\ $$$$ \\ $$$${t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}={a}\left({t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}\right)+\left({bt}^{\mathrm{2}} +{c}\right)\left(\mathrm{3}−{t}^{\mathrm{2}} \right) \\ $$$${t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}={at}^{\mathrm{4}} −\mathrm{6}{at}^{\mathrm{2}} +{a}+\mathrm{3}{bt}^{\mathrm{2}} −{bt}^{\mathrm{4}} +\mathrm{3}{c}−{ct}^{\mathrm{2}} \\ $$$${t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}={t}^{\mathrm{4}} \left({a}−{b}\right)+{t}^{\mathrm{2}} \left(−\mathrm{6}{a}+\mathrm{3}{b}−{c}\right)+\left({a}+{c}\right) \\ $$$${a}−{b}=\mathrm{1} \\ $$$$−\mathrm{6}{a}+\mathrm{3}{b}−{c}=\mathrm{2} \\ $$$${a}+{c}=\mathrm{1} \\ $$$$−\mathrm{6}{a}+\mathrm{3}\left({a}−\mathrm{1}\right)−\left(\mathrm{1}−{a}\right)=\mathrm{2} \\ $$$$−\mathrm{6}{a}+\mathrm{3}{a}+{a}−\mathrm{3}−\mathrm{1}=\mathrm{2} \\ $$$$−\mathrm{2}{a}=\mathrm{6}\:\:{a}=−\mathrm{3} \\ $$$${b}={a}−\mathrm{1}=−\mathrm{4} \\ $$$${c}=\mathrm{1}−{a}=\mathrm{4} \\ $$$${a}=−\mathrm{3}\:\:\:{b}=−\mathrm{4}\:\:\:{c}=\mathrm{4} \\ $$$$\mathrm{2}\int\frac{{a}\left({t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}\right)+\left({bt}^{\mathrm{2}} +{c}\right)\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{\left({t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\mathrm{2}\int\frac{−\mathrm{3}}{\mathrm{3}−{t}^{\mathrm{2}} }{dt}+\mathrm{2}\int\frac{\left(−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}\right)}{{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$−\mathrm{6}\int\frac{{dt}}{\mathrm{3}−{t}^{\mathrm{2}} }−\mathrm{8}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{6}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{3}}−\mathrm{8}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)−\mathrm{6}} \\ $$$$\mathrm{6}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{3}}−\mathrm{8}\int\frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{8}} \\ $$$$\mathrm{6}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\left(\frac{{t}−\sqrt{\mathrm{3}}}{{t}+\sqrt{\mathrm{3}}}\right)−\mathrm{8}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{8}}\:}{ln}\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{8}}}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{8}}}\right)+{c} \\ $$$$=\sqrt{\mathrm{3}}\:{ln}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}−\sqrt{\mathrm{3}}}{{tan}\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{3}}}\right)−\sqrt{\mathrm{2}}\:{ln}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{{tan}\frac{{x}}{\mathrm{2}}}−\mathrm{2}\sqrt{\mathrm{2}}}{{tan}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{{tan}\frac{{x}}{\mathrm{2}}}+\mathrm{2}\sqrt{\mathrm{2}}}\right)+{c} \\ $$$$ \\ $$