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Question Number 214341 by liuxinnan last updated on 06/Dec/24
∫(dx/(3+cosx))=?
$$\int\frac{{dx}}{\mathrm{3}+{cosx}}=? \\ $$
Answered by chhaythean last updated on 06/Dec/24
let, t = tan(x/2) ⇒ dt = (1/2)sec^2 ((x/2))dx or dt = (1/2)(1+tan^2 ((x/2)))dx = (1/2)(1+t^2 )dx ⇒ cosx =((1−t^2 )/(1+t^2 )) ⇒ ∫(dx/(3+cosx)) =2 ∫(1/((1+t)^2 ))×(1/(3+((1−t^2 )/(1+t^2 ))))dt = 2∫(1/((1+t^2 )))×(1/((3+3t^2 +1−t^2 )/(1+t^2 )))dt =2∫(dt/(4+2t^2 )) = ∫(dt/(2+t^2 )) = ∫(dt/( ((√2))^2 +t^2 )) =(1/( (√2))) arctan((t/( (√2))))+C =((√2)/2)arctan(((tan((x/2)))/( (√2))))+C
$$\mathrm{let},\:\mathrm{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}}\:\Rightarrow\:\mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\mathrm{d}{x} \\ $$$$\mathrm{or}\:\mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\mathrm{d}{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{d}{x} \\ $$$$\Rightarrow\:\mathrm{cos}{x}\:=\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\: \\ $$$$\Rightarrow\:\int\frac{\mathrm{d}{x}}{\mathrm{3}+\mathrm{cos}{x}}\:=\mathrm{2}\:\int\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\mathrm{dt} \\ $$$$=\:\mathrm{2}\int\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}×\frac{\mathrm{1}}{\frac{\mathrm{3}+\mathrm{3t}^{\mathrm{2}} +\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\mathrm{dt}\: \\ $$$$=\mathrm{2}\int\frac{\mathrm{dt}}{\mathrm{4}+\mathrm{2t}^{\mathrm{2}} }\:=\:\int\frac{\mathrm{dt}}{\mathrm{2}+\mathrm{t}^{\mathrm{2}} }\:=\:\int\frac{\mathrm{dt}}{\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{arctan}\left(\frac{\mathrm{t}}{\:\sqrt{\mathrm{2}}}\right)+{C} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\left(\frac{\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}}}\right)+{C} \\ $$
Answered by shunmisaki007 last updated on 06/Dec/24
Let t=tan((x/2)) ; cos((x/2))=(1/( (√(1+t^2 )))), sin((x/2))=(t/( (√(1+t^2 )))), cos(x)=cos^2 ((x/2))−sin^2 ((x/2))=((1−t^2 )/(1+t^2 )), dt=(1/2)sec^2 ((x/2))dx=((1+t^2 )/2)dx or dx=(2/(1+t^2 ))dt. So ∫(dx/(3+cos(x)))=∫((2/(1+t^2 ))/(3+((1−t^2 )/(1+t^2 ))))dt=∫(2/(4+2t^2 ))dt=∫(1/(2+t^2 ))dt Let t=(√2)tan(u) ; dt=(√2)sec^2 (u)du. So ∫(1/(2+t^2 ))dt=∫(((√2)sec^2 (u))/(2+2tan^2 (u)))du=((√2)/2)∫du =((√2)/2)u+C=((√2)/2)tan^(−1) ((((√2)t)/2))+C =((√2)/2)tan^(−1) ((((√2)tan((x/2)))/2))+C ∴∫(dx/(3+cos(x)))=((√2)/2)tan^(−1) ((((√2)tan((x/2)))/2))+C ★
$$\mathrm{Let}\:{t}=\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$;\:\mathrm{cos}\left(\frac{{x}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }},\:\mathrm{sin}\left(\frac{{x}}{\mathrm{2}}\right)=\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}, \\ $$$$\mathrm{cos}\left({x}\right)=\mathrm{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }, \\ $$$${dt}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){dx}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}}{dx}\:\mathrm{or}\:{dx}=\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}. \\ $$$$\mathrm{So}\:\int\frac{{dx}}{\mathrm{3}+\mathrm{cos}\left({x}\right)}=\int\frac{\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{3}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{dt}=\int\frac{\mathrm{2}}{\mathrm{4}+\mathrm{2}{t}^{\mathrm{2}} }{dt}=\int\frac{\mathrm{1}}{\mathrm{2}+{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{Let}\:{t}=\sqrt{\mathrm{2}}\mathrm{tan}\left({u}\right) \\ $$$$;\:{dt}=\sqrt{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \left({u}\right){du}. \\ $$$$\mathrm{So}\:\int\frac{\mathrm{1}}{\mathrm{2}+{t}^{\mathrm{2}} }{dt}=\int\frac{\sqrt{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \left({u}\right)}{\mathrm{2}+\mathrm{2tan}^{\mathrm{2}} \left({u}\right)}{du}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int{du} \\ $$$$\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}+{C}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}{t}}{\mathrm{2}}\right)+{C} \\ $$$$\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}}\right)+{C} \\ $$$$\therefore\int\frac{{dx}}{\mathrm{3}+\mathrm{cos}\left({x}\right)}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}}\right)+{C}\:\bigstar \\ $$$$ \\ $$

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