Question Number 27186 by abdo imad last updated on 02/Jan/18
$${find}\:{I}=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{{cosx}\:+\mathrm{2}{sinx}}\:. \\ $$
Commented by abdo imad last updated on 04/Jan/18
![we do the changement tan((x/2))=t I= ∫_0 ^∞ (((2dt)/(1+t^2 ))/(((1−t^2 )/(1+t^2 ))+((4t)/(1+t^2 )))) = ∫_0 ^∞ ((2dt)/(−t^2 +4t +1)) = −2 ∫_0 ^∞ (dt/(t^2 −4t −1)) = −2 ∫_0 ^∞ (dt/(t^2 −4t +4 −5)) =−2∫_0 ^∞ (dt/((t−2)^2 −5))=−2∫_0 ^∞ (dt/((t−2+(√5))(t−2−(√(5))))) = (1/( (√5)))∫_0 ^∞ ((1/(t−2+(√)5)) −(1/(t−2−(√5)))) =(1/( (√5))) [ ln/((t−2+(√5))/(t−2−(√5)))/]_0 ^∝ =(1/( (√5)))(−ln/((−2+(√5))/(−2−(√5)))/)= (1/( (√5))) ln(((2+(√5))/(−2+(√5)))) .](https://www.tinkutara.com/question/Q27257.png)
$${we}\:{do}\:{the}\:{changement}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{dt}}{−{t}^{\mathrm{2}} +\mathrm{4}{t}\:+\mathrm{1}} \\ $$$$=\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{4}{t}\:−\mathrm{1}}\:=\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} \:−\mathrm{4}{t}\:+\mathrm{4}\:−\mathrm{5}} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({t}−\mathrm{2}\right)^{\mathrm{2}} \:−\mathrm{5}}=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({t}−\mathrm{2}+\sqrt{\mathrm{5}}\right)\left({t}−\mathrm{2}−\sqrt{\left.\mathrm{5}\right)}\right.} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{t}−\mathrm{2}+\sqrt{}\mathrm{5}}\:−\frac{\mathrm{1}}{{t}−\mathrm{2}−\sqrt{\mathrm{5}}}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\left[\:{ln}/\frac{{t}−\mathrm{2}+\sqrt{\mathrm{5}}}{{t}−\mathrm{2}−\sqrt{\mathrm{5}}}/\right]_{\mathrm{0}} ^{\propto} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(−{ln}/\frac{−\mathrm{2}+\sqrt{\mathrm{5}}}{−\mathrm{2}−\sqrt{\mathrm{5}}}/\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{5}}}{−\mathrm{2}+\sqrt{\mathrm{5}}}\right)\:\:. \\ $$
Answered by mrW1 last updated on 03/Jan/18
$${I}=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{{cosx}\:+\mathrm{2}{sinx}}\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{cosx}\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}{sinx}}\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\mathrm{sin}\:\alpha\:{cosx}\:+\mathrm{cos}\:\alpha\:{sinx}}\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{{sin}\left({x}+\alpha\right)}\: \\ $$$$…… \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\mathrm{ln}\:\left(\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}}\right) \\ $$