Question Number 68874 by mathmax by abdo last updated on 16/Sep/19
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\mathrm{1}+{cosx}\:+\mathrm{3}{sinx}} \\ $$
Commented by mathmax by abdo last updated on 24/Sep/19
![let I =∫_0 ^(2π) (dx/(1+cosx +3sinx)) ⇒I =∫_0 ^π (dx/(1+cosx +3sinx)) +∫_π ^(2π) (dx/(1+cosx +3sinx)) =H +K H =_(tan((x/2))=t) ∫_0 ^∞ ((2dt)/((1+t^2 )(1+((1−t^2 )/(1+t^2 ))+3((2t)/(1+t^2 ))))) =∫_0 ^∞ ((2dt)/(1+t^2 +1−t^2 +6t)) =∫_0 ^∞ ((2dt)/(6t +2)) =∫_0 ^∞ (dt/(3t+1)) =∞ K =_(x =π+u) ∫_0 ^π (du/(1−cosu−3sinu)) =_(tan((u/2))=α) ∫_0 ^∞ ((2dα)/((1+α^2 )(1−((1−α^2 )/(1+α^2 ))−3((2α)/(1+α^2 ))))) =∫_0 ^∞ ((2dα)/(1+α^2 −1+α^2 −6α)) =∫_0 ^∞ ((2dα)/(2α^2 −6α)) =∫_0 ^∞ (dα/(α^2 −3α)) =−(1/3)∫_0 ^∞ ((1/α)−(1/(α−3)))dα =−(1/3)[ln∣(α/(α−3))∣]_0 ^(+∞) =−(1/3)(−ln((1/3))) =−((ln3)/3) we see that I is divergent...!](https://www.tinkutara.com/question/Q69484.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{\mathrm{1}+{cosx}\:+\mathrm{3}{sinx}}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{1}+{cosx}\:+\mathrm{3}{sinx}} \\ $$$$+\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{dx}}{\mathrm{1}+{cosx}\:+\mathrm{3}{sinx}}\:={H}\:+{K} \\ $$$${H}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{6}{t}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{6}{t}\:+\mathrm{2}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\mathrm{3}{t}+\mathrm{1}}\:=\infty \\ $$$${K}\:=_{{x}\:=\pi+{u}} \:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{du}}{\mathrm{1}−{cosu}−\mathrm{3}{sinu}}\:=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)=\alpha} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{d}\alpha}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\mathrm{1}−\alpha^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{2}} }−\mathrm{3}\frac{\mathrm{2}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{6}\alpha}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{d}\alpha}{\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{6}\alpha}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{d}\alpha}{\alpha^{\mathrm{2}} −\mathrm{3}\alpha} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\:\left(\frac{\mathrm{1}}{\alpha}−\frac{\mathrm{1}}{\alpha−\mathrm{3}}\right){d}\alpha\:=−\frac{\mathrm{1}}{\mathrm{3}}\left[{ln}\mid\frac{\alpha}{\alpha−\mathrm{3}}\mid\right]_{\mathrm{0}} ^{+\infty} =−\frac{\mathrm{1}}{\mathrm{3}}\left(−{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$=−\frac{{ln}\mathrm{3}}{\mathrm{3}}\:\:\:{we}\:{see}\:{that}\:{I}\:{is}\:{divergent}…! \\ $$