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Question Number 137192 by bobhans last updated on 30/Mar/21
∫ ((cos x)/(1+cos x+sin x)) dx =?
$$\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:=? \\ $$
Answered by bemath last updated on 30/Mar/21
Answered by mathmax by abdo last updated on 30/Mar/21
I =∫ ((cosx)/(1+cosx+sinx))dx we do the changement tan((x/2))=t ⇒ I =∫ (((1−t^2 )/(1+t^2 ))/(1+((1−t^2 )/(1+t^2 )) +((2t)/(1+t^2 ))))((2dt)/(1+t^2 )) =2∫ ((1−t^2 )/((1+t^2 )(1+t^2 +1−t^2 +2t)))dt =2 ∫ ((1−t^2 )/((1+t^2 )(2+2t))) =2 ∫ (((1−t)(1+t))/((t^2 +1)(t+1))) =2 ∫((1−t)/(1+t^2 ))dt =2arctant−ln(1+t^2 ) +C =2.(x/2)−ln(1+tan^2 ((t/2))) +C =x−ln(1+tan^2 ((x/2))) +C
$$\mathrm{I}\:=\int\:\frac{\mathrm{cosx}}{\mathrm{1}+\mathrm{cosx}+\mathrm{sinx}}\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int\:\:\frac{\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:+\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\frac{\mathrm{2dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:=\mathrm{2}\int\:\:\:\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}\right)}\mathrm{dt} \\ $$$$=\mathrm{2}\:\int\:\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{2}+\mathrm{2t}\right)}\:=\mathrm{2}\:\int\:\:\frac{\left(\mathrm{1}−\mathrm{t}\right)\left(\mathrm{1}+\mathrm{t}\right)}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{t}+\mathrm{1}\right)}\:=\mathrm{2}\:\int\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\mathrm{2arctant}−\mathrm{ln}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\:+\mathrm{C} \\ $$$$=\mathrm{2}.\frac{\mathrm{x}}{\mathrm{2}}−\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{t}}{\mathrm{2}}\right)\right)\:+\mathrm{C}\:=\mathrm{x}−\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)\:+\mathrm{C} \\ $$

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