Question Number 53112 by maxmathsup by imad last updated on 21/Jan/19
![calculate ∫_0 ^π ((1+2sinx)/(3 +2cosx))dx let A =∫_0 ^π ((1+2sinx)/(3 +2cosx))dx changement tan((x/2))=t give A =∫_0 ^∞ ((1+((4t)/(1+t^2 )))/(3+2((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =2 ∫_0 ^∞ ((1+t^2 +4t)/((1+t^2 )^2 (((3+3t^2 +2−2t^2 )/(1+t^2 )))))dt =2 ∫_0 ^∞ ((t^2 +4t +1)/((1+t^2 )(5+t^2 )))dt let decompose F(t)=((t^2 +4t+1)/((t^2 +1)(t^2 +5))) F(t)=((at +b)/(t^2 +1)) +((ct +d)/(t^2 +5)) ⇒(at+b)(t^2 +5)+(ct+d)(t^2 +1) =t^2 +4t +1 ⇒ at^3 +5at +bt^2 +5b +ct^3 +ct +dt^2 +d =t^2 +4t +1 ⇒ (a+c)t^3 +(b+d)t^2 +(5a+c)t +5b +d =t^2 +4t +1 ⇒a+c=0 and b+d=1 and 5a+c =4 and 5b+d =1 ⇒c=−a ⇒a=1 ⇒c=−1 we have d=1−b ⇒5b +1−b =1 ⇒b=0 ⇒d=1 ⇒ F(t)=(t/(t^2 +1)) +((−t +1)/(t^2 +5)) ⇒ A =2 ∫_0 ^∞ F(t)dt =∫_0 ^∞ ((2t)/(t^2 +1))dt +∫_0 ^∞ ((−2t +2)/(t^2 +5))dt =[ln(((t^2 +1)/(t^2 +5)))]_0 ^(+∞) +2 ∫_0 ^∞ (dt/(t^2 +5)) =ln(5) + 2 ∫_0 ^∞ (dt/(t^2 +5)) but ∫_0 ^∞ (dt/(t^2 +5))dt =_(t =(√5)u ) ∫_0 ^∞ (((√5)du)/(5(1+u^2 ))) =(1/(√5)) [artanu]_0 ^(+∞) =(π/(2(√5))) ⇒ A =ln(5) +(π/(2(√5))) .](Q53112.png)
$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{1}+\mathrm{2}{sinx}}{\mathrm{3}\:+\mathrm{2}{cosx}}{dx} \\ $$$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}+\mathrm{2}{sinx}}{\mathrm{3}\:+\mathrm{2}{cosx}}{dx}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}+\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{3}+\mathrm{2}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\frac{\mathrm{3}+\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)}{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} +\mathrm{4}{t}\:+\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{5}+{t}^{\mathrm{2}} \right)}{dt}\:\:{let}\:{decompose}\:{F}\left({t}\right)=\frac{{t}^{\mathrm{2}} \:+\mathrm{4}{t}+\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{5}\right)} \\ $$$${F}\left({t}\right)=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\:\Rightarrow\left({at}+{b}\right)\left({t}^{\mathrm{2}} \:+\mathrm{5}\right)+\left({ct}+{d}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:+\mathrm{1}\:\Rightarrow \\ $$$${at}^{\mathrm{3}} \:+\mathrm{5}{at}\:+{bt}^{\mathrm{2}} \:+\mathrm{5}{b}\:+{ct}^{\mathrm{3}} \:+{ct}\:+{dt}^{\mathrm{2}} \:+{d}\:={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:+\mathrm{1}\:\Rightarrow \\ $$$$\left({a}+{c}\right){t}^{\mathrm{3}} \:+\left({b}+{d}\right){t}^{\mathrm{2}} \:+\left(\mathrm{5}{a}+{c}\right){t}\:+\mathrm{5}{b}\:+{d}\:={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:+\mathrm{1}\:\Rightarrow{a}+{c}=\mathrm{0}\:{and}\:{b}+{d}=\mathrm{1}\:{and} \\ $$$$\mathrm{5}{a}+{c}\:=\mathrm{4}\:{and}\:\mathrm{5}{b}+{d}\:=\mathrm{1}\:\Rightarrow{c}=−{a}\:\Rightarrow{a}=\mathrm{1}\:\Rightarrow{c}=−\mathrm{1}\: \\ $$$${we}\:{have}\:{d}=\mathrm{1}−{b}\:\Rightarrow\mathrm{5}{b}\:+\mathrm{1}−{b}\:=\mathrm{1}\:\Rightarrow{b}=\mathrm{0}\:\Rightarrow{d}=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{t}\:+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\:\:\Rightarrow\:{A}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{F}\left({t}\right){dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:+\int_{\mathrm{0}} ^{\infty} \:\frac{−\mathrm{2}{t}\:+\mathrm{2}}{{t}^{\mathrm{2}} \:+\mathrm{5}}{dt} \\ $$$$=\left[{ln}\left(\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\right)\right]_{\mathrm{0}} ^{+\infty} \:\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\:={ln}\left(\mathrm{5}\right)\:+\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{5}}{dt}\:=_{{t}\:=\sqrt{\mathrm{5}}{u}\:} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\sqrt{\mathrm{5}}{du}}{\mathrm{5}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\:\left[{artanu}\right]_{\mathrm{0}} ^{+\infty} \:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$${A}\:={ln}\left(\mathrm{5}\right)\:+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{5}}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
![∫(1/(3+2cosx))dx+∫((2sinx)/(3+2cosx))dx ∫((1+tan^2 (x/2))/(3+3tan^2 (x/2)+2−2tan^2 (x/2)))dx−∫((d(3+2cosx))/(3+2cosx)) ∫((sec^2 (x/2))/(5+tan^2 (x/2)))dx−ln(3+2cosx)+c 2∫((d(tan(x/2)))/(5+tan^2 (x/2)))−do ∣2×(1/(√5))tan^(−1) (((tan(x/2))/(√5)))−ln(3+2cosx)∣_0 ^π [{(2/(√5))tan^(−1) (((tan(π/2))/(√5)))−ln(3+2cosπ)}−{(2/(√5))tan^(−1) (((tan0)/(√5)))−ln(3+2cos0)}] =[{(2/(√5))((π/2))−ln(3−2)}−{(2/(√5))tan^(−1) (0)−ln5}] =(π/(√5))+ln5](Q53203.png)
$$\int\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}{cosx}}{dx}+\int\frac{\mathrm{2}{sinx}}{\mathrm{3}+\mathrm{2}{cosx}}{dx} \\ $$$$\int\frac{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{3}+\mathrm{3}{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}−\mathrm{2}{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}−\int\frac{{d}\left(\mathrm{3}+\mathrm{2}{cosx}\right)}{\mathrm{3}+\mathrm{2}{cosx}} \\ $$$$\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{5}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}−{ln}\left(\mathrm{3}+\mathrm{2}{cosx}\right)+{c} \\ $$$$\mathrm{2}\int\frac{{d}\left({tan}\frac{{x}}{\mathrm{2}}\right)}{\mathrm{5}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}−{do} \\ $$$$\mid\mathrm{2}×\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}{tan}^{−\mathrm{1}} \left(\frac{{tan}\frac{{x}}{\mathrm{2}}}{\sqrt{\mathrm{5}}}\right)−{ln}\left(\mathrm{3}+\mathrm{2}{cosx}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$$\left[\left\{\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}{tan}^{−\mathrm{1}} \left(\frac{{tan}\frac{\pi}{\mathrm{2}}}{\sqrt{\mathrm{5}}}\right)−{ln}\left(\mathrm{3}+\mathrm{2}{cos}\pi\right)\right\}−\left\{\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}{tan}^{−\mathrm{1}} \left(\frac{{tan}\mathrm{0}}{\sqrt{\mathrm{5}}}\right)−{ln}\left(\mathrm{3}+\mathrm{2}{cos}\mathrm{0}\right)\right\}\right] \\ $$$$=\left[\left\{\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\left(\frac{\pi}{\mathrm{2}}\right)−{ln}\left(\mathrm{3}−\mathrm{2}\right)\right\}−\left\{\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}{tan}^{−\mathrm{1}} \left(\mathrm{0}\right)−{ln}\mathrm{5}\right\}\right] \\ $$$$=\frac{\pi}{\sqrt{\mathrm{5}}}+{ln}\mathrm{5} \\ $$