Question Number 31070 by abdo imad last updated on 02/Mar/18
$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+\mathrm{2}{cosx}}\:. \\ $$
Commented by abdo imad last updated on 03/Mar/18
![the ch.tan((x/2))=t give I= ∫_0 ^(+∞) (1/(1+2((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^(+∞) ((2dt)/(1+t^2 +2(1−t^2 ))) = ∫_0 ^(+∞) ((2dt)/(3−t^2 )) =(1/(2(√3)))∫_0 ^(+∞) 2( (1/( (√3) −t)) +(1/( (√3) +t)))dt =(1/( (√3)))[ln((√3) +t)−ln((√3) −t)]_0 ^(+∞) =(1/( (√3)))[ln(∣(((√3) +t)/( (√3) −t)))∣]_0 ^(+∞) ⇒ I=0](https://www.tinkutara.com/question/Q31203.png)
$${the}\:{ch}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}=\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{+\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$$=\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{3}−{t}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{+\infty} \mathrm{2}\left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:−{t}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+{t}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[{ln}\left(\sqrt{\mathrm{3}}\:+{t}\right)−{ln}\left(\sqrt{\mathrm{3}}\:−{t}\right)\right]_{\mathrm{0}} ^{+\infty} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[{ln}\left(\mid\frac{\sqrt{\mathrm{3}}\:+{t}}{\:\sqrt{\mathrm{3}}\:−{t}}\right)\mid\right]_{\mathrm{0}} ^{+\infty} \Rightarrow \\ $$$${I}=\mathrm{0} \\ $$