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Question Number 53080 by Abdo msup. last updated on 17/Jan/19
calculate ∫_0 ^π ((cos^2 x)/(2+3sin(2x)))dx
$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cos}^{\mathrm{2}} {x}}{\mathrm{2}+\mathrm{3}{sin}\left(\mathrm{2}{x}\right)}{dx} \\ $$
Commented by maxmathsup by imad last updated on 17/Jan/19
let A =∫_0 ^π ((cos^2 x)/(2+3sin(2x)))dx ⇒A=∫_0 ^π ((1+cos(2x))/(2(2+3sin(2x))))dx =_(2x=t) ∫_0 ^(2π) ((1+cos(t))/(2(2+3sin(t)))) (dt/2) =(1/4) ∫_0 ^(2π) ((1+cos(t))/(2+3sin(t)))dt ⇒ 4A =∫_0 ^(2π) ((1+cost)/(2+3sint)) dt = Re( ∫_0 ^(2π) ((1+e^(it) )/(2+3sint))dt) let changement e^(it) =z give ∫_0 ^(2π) ((1+e^(it) )/(2+3sint))dt =∫_(∣z∣=1) ((1+z)/(2+3((z−z^(−1) )/(2i))))(dz/(iz)) = ∫_(∣z∣=1) ((2i(1+z))/(iz(4i +3z−3z^(−1) )))dz =∫_(∣z∣=1) ((2(1+z))/(4iz +3z^2 −3))dz let consider the complex function ϕ(z)=((2(1+z))/(3z^2 +4iz −3)) poles of ϕ? let find roots of 3z^2 +4iz −3 Δ^′ =(2i)^2 −3(−3) =−4+9 =5 ⇒z_1 =((−2i+(√5))/3) and z_2 =((−2i−(√5))/3)(are poles of ϕ) ∣z_1 ∣−1 =(1/3)(√(4+5))=3 >0 ⇒z_1 is out of circle ⇒Res(ϕ,z_1 )=0 ∣z_2 ∣−1 =(1/3)(√(4+5))=3>0 ⇒z_2 is out of circle ⇒Res(ϕ,z_2 )=0 ∫_(∣z∣=1) ϕ(z)dz =2iπ{ Res(ϕ,z_1 ) +Res(ϕ,z_2 )}=0 ⇒Re(∫ ϕ(zdz)=0 ⇒ A =0 .
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cos}^{\mathrm{2}} {x}}{\mathrm{2}+\mathrm{3}{sin}\left(\mathrm{2}{x}\right)}{dx}\:\Rightarrow{A}=\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}\left(\mathrm{2}+\mathrm{3}{sin}\left(\mathrm{2}{x}\right)\right)}{dx} \\ $$$$=_{\mathrm{2}{x}={t}} \:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{1}+{cos}\left({t}\right)}{\mathrm{2}\left(\mathrm{2}+\mathrm{3}{sin}\left({t}\right)\right)}\:\frac{{dt}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\mathrm{1}+{cos}\left({t}\right)}{\mathrm{2}+\mathrm{3}{sin}\left({t}\right)}{dt}\:\Rightarrow \\ $$$$\mathrm{4}{A}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\mathrm{1}+{cost}}{\mathrm{2}+\mathrm{3}{sint}}\:{dt}\:\:=\:{Re}\left(\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{1}+{e}^{{it}} }{\mathrm{2}+\mathrm{3}{sint}}{dt}\right)\:{let}\:\:{changement}\:{e}^{{it}} ={z}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{1}+{e}^{{it}} }{\mathrm{2}+\mathrm{3}{sint}}{dt}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{1}+{z}}{\mathrm{2}+\mathrm{3}\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}\frac{{dz}}{{iz}}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{i}\left(\mathrm{1}+{z}\right)}{{iz}\left(\mathrm{4}{i}\:+\mathrm{3}{z}−\mathrm{3}{z}^{−\mathrm{1}} \right)}{dz} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}\left(\mathrm{1}+{z}\right)}{\mathrm{4}{iz}\:+\mathrm{3}{z}^{\mathrm{2}} \:−\mathrm{3}}{dz}\:\:{let}\:{consider}\:{the}\:{complex}\:{function}\: \\ $$$$\varphi\left({z}\right)=\frac{\mathrm{2}\left(\mathrm{1}+{z}\right)}{\mathrm{3}{z}^{\mathrm{2}} \:+\mathrm{4}{iz}\:−\mathrm{3}}\:\:{poles}\:{of}\:\varphi?\:\:{let}\:{find}\:{roots}\:{of}\:\mathrm{3}{z}^{\mathrm{2}} \:+\mathrm{4}{iz}\:−\mathrm{3} \\ $$$$\Delta^{'} =\left(\mathrm{2}{i}\right)^{\mathrm{2}} −\mathrm{3}\left(−\mathrm{3}\right)\:=−\mathrm{4}+\mathrm{9}\:=\mathrm{5}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{2}{i}+\sqrt{\mathrm{5}}}{\mathrm{3}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{2}{i}−\sqrt{\mathrm{5}}}{\mathrm{3}}\left({are}\:{poles}\:{of}\:\varphi\right) \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{4}+\mathrm{5}}=\mathrm{3}\:>\mathrm{0}\:\Rightarrow{z}_{\mathrm{1}} \:{is}\:{out}\:{of}\:{circle}\:\Rightarrow{Res}\left(\varphi,{z}_{\mathrm{1}} \right)=\mathrm{0} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{4}+\mathrm{5}}=\mathrm{3}>\mathrm{0}\:\Rightarrow{z}_{\mathrm{2}} {is}\:{out}\:{of}\:{circle}\:\Rightarrow{Res}\left(\varphi,{z}_{\mathrm{2}} \right)=\mathrm{0}\: \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:+{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\right\}=\mathrm{0}\:\Rightarrow{Re}\left(\int\:\varphi\left({zdz}\right)=\mathrm{0}\:\Rightarrow\right. \\ $$$${A}\:=\mathrm{0}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19
∫(((1+cos2x)/2)/(2+3sin2x))dx (1/2)∫((1+cos2x)/(2+3sin2x))dx (1/2)∫(dx/(2+3(((2tanx)/(1+tan^2 x)))))+(1/(12))∫((d(2+3sin2x))/(2+3sin2x)) (1/2)∫((sec^2 xdx)/(2+2tan^2 x+6tanx))+(1/(12))ln(2+3sin2x)+c (1/4)∫((d(tanx))/(tan^2 x+3tanx+1))+do (1/4)∫((d(tanx))/(tan^2 x+2tanx×(3/2)+(9/4)+1−(9/4)))+do (1/4)∫((d(tanx))/((tanx+(3/2))^2 −((√(5/4)) )^2 ))+do (1/4)∫((d(tanx+(3/2)))/((tanx+(3/2))^2 −((√(5/4)) )^2 ))+do (1/4)×(1/(2(((√5)/2))))ln(((tanx+(3/2)−((√5)/2))/(tanx+(3/2)+((√5)/2))))+(1/(12))ln(2+3sin2x)+c now we have to put limit.. (1/(4(√5)))∣ln(((tanx+(3/2)−((√5)/2))/(tanx+(3/2)+((√5)/2))))+(1/(12))ln(2+3sin2x)∣_0 ^π answer is zero...
$$\int\frac{\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}{\mathrm{2}+\mathrm{3}{sin}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}+\mathrm{3}{sin}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{2}+\mathrm{3}\left(\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}\right)}+\frac{\mathrm{1}}{\mathrm{12}}\int\frac{{d}\left(\mathrm{2}+\mathrm{3}{sin}\mathrm{2}{x}\right)}{\mathrm{2}+\mathrm{3}{sin}\mathrm{2}{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{sec}^{\mathrm{2}} {xdx}}{\mathrm{2}+\mathrm{2}{tan}^{\mathrm{2}} {x}+\mathrm{6}{tanx}}+\frac{\mathrm{1}}{\mathrm{12}}{ln}\left(\mathrm{2}+\mathrm{3}{sin}\mathrm{2}{x}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left({tanx}\right)}{{tan}^{\mathrm{2}} {x}+\mathrm{3}{tanx}+\mathrm{1}}+{do} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left({tanx}\right)}{{tan}^{\mathrm{2}} {x}+\mathrm{2}{tanx}×\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{1}−\frac{\mathrm{9}}{\mathrm{4}}}+{do} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left({tanx}\right)}{\left({tanx}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}\:\right)^{\mathrm{2}} }+{do} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left({tanx}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{\left({tanx}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}\:\right)^{\mathrm{2}} }+{do} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)}{ln}\left(\frac{{tanx}+\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{{tanx}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{12}}{ln}\left(\mathrm{2}+\mathrm{3}{sin}\mathrm{2}{x}\right)+{c} \\ $$$${now}\:{we}\:{have}\:{to}\:{put}\:{limit}.. \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{5}}}\mid{ln}\left(\frac{{tanx}+\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{{tanx}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{12}}{ln}\left(\mathrm{2}+\mathrm{3}{sin}\mathrm{2}{x}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$${answer}\:{is}\:{zero}… \\ $$$$ \\ $$$$ \\ $$

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