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Question Number 63927 by aliesam last updated on 11/Jul/19

∫_0 ^π (dx/((3+2cos x)^2 ))

$$\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\left(\mathrm{3}+\mathrm{2}{cos}\:{x}\right)^{\mathrm{2}} } \\ $$

Commented byaliesam last updated on 12/Jul/19

god bless you sir ..well done..

$${god}\:{bless}\:{you}\:{sir}\:..{well}\:{done}.. \\ $$

Commented bymathmax by abdo last updated on 12/Jul/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

Commented byaliesam last updated on 11/Jul/19

thank you sir

$${thank}\:{you}\:{sir}\: \\ $$

Commented byaliesam last updated on 11/Jul/19

but i think that the solution is ((3π)/(5(√5)))

$${but}\:{i}\:{think}\:{that}\:{the}\:{solution}\:{is}\:\frac{\mathrm{3}\pi}{\mathrm{5}\sqrt{\mathrm{5}}} \\ $$

Commented bymathmax by abdo last updated on 12/Jul/19

let f(t)=∫_0 ^π (dx/(t+2cosx)) ⇒f^′ (t) =−∫_0 ^π (dx/((t+2cosx)^2 )) ⇒ ∫_0 ^π (dx/((t+2cosx)^2 )) =−f^′ (t) and ∫_0 ^π (dx/((3+2cosx)^2 )) =−f^′ (3) let calculate f(t) changement tan((x/2))=u give f(t)=∫_0 ^∞ ((2du)/((1+u^2 )( t +2((1−u^2 )/(1+u^2 ))))) =∫_0 ^∞ ((2du)/(t+tu^2 +2−2u^2 )) =∫_0 ^∞ ((2du)/((t−2)u^2 +t+2)) due to our case we take t>2 ⇒ f(t) =(2/((t−2)))∫_0 ^∞ (du/(u^2 +((t+2)/(t−2)))) =_(u=(√((t+2)/(t−2)))α) (2/(t−2)) ∫_0 ^∞ (1/(((t+2)/(t−2))(1+α^2 )))(√((t+2)/(t−2)))dα =(2/(√(t^2 −4))) (π/2) =(π/(√(t^2 −4))) ⇒f^′ (t) =π{(t^2 −4)^(−(1/2)) }^((1)) =−(π/2)(2t)(t^2 −4)^(−(3/2)) =((−πt)/((t^2 −4)(√(t^2 −4)))) ⇒ ∫_0 ^π (dx/((t+2cosx)^2 )) =((πt)/((t^2 −4)(√(t^2 −4)))) and ∫_0 ^π (dx/((3+2cosx)^2 )) =((3π)/((9−4)(√(9−4)))) =((3π)/(5(√5))) .

$${let}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{{t}+\mathrm{2}{cosx}}\:\Rightarrow{f}^{'} \left({t}\right)\:=−\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\left({t}+\mathrm{2}{cosx}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\left({t}+\mathrm{2}{cosx}\right)^{\mathrm{2}} }\:=−{f}^{'} \left({t}\right)\:{and}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\left(\mathrm{3}+\mathrm{2}{cosx}\right)^{\mathrm{2}} }\:=−{f}^{'} \left(\mathrm{3}\right) \\ $$ $${let}\:{calculate}\:{f}\left({t}\right)\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}\:{give} \\ $$ $${f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\:{t}\:+\mathrm{2}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{du}}{{t}+{tu}^{\mathrm{2}} \:+\mathrm{2}−\mathrm{2}{u}^{\mathrm{2}} } \\ $$ $$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{du}}{\left({t}−\mathrm{2}\right){u}^{\mathrm{2}} \:+{t}+\mathrm{2}}\:\:{due}\:{to}\:{our}\:{case}\:{we}\:{take}\:{t}>\mathrm{2}\:\Rightarrow \\ $$ $${f}\left({t}\right)\:=\frac{\mathrm{2}}{\left({t}−\mathrm{2}\right)}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\frac{{t}+\mathrm{2}}{{t}−\mathrm{2}}}\:=_{{u}=\sqrt{\frac{{t}+\mathrm{2}}{{t}−\mathrm{2}}}\alpha} \:\:\frac{\mathrm{2}}{{t}−\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\frac{{t}+\mathrm{2}}{{t}−\mathrm{2}}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\sqrt{\frac{{t}+\mathrm{2}}{{t}−\mathrm{2}}}{d}\alpha \\ $$ $$=\frac{\mathrm{2}}{\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}\:\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\sqrt{{t}^{\mathrm{2}} \:−\mathrm{4}}}\:\Rightarrow{f}^{'} \left({t}\right)\:=\pi\left\{\left({t}^{\mathrm{2}} −\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right\}^{\left(\mathrm{1}\right)} \\ $$ $$=−\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{t}\right)\left({t}^{\mathrm{2}} −\mathrm{4}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:=\frac{−\pi{t}}{\left({t}^{\mathrm{2}} −\mathrm{4}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\left({t}+\mathrm{2}{cosx}\right)^{\mathrm{2}} }\:=\frac{\pi{t}}{\left({t}^{\mathrm{2}} −\mathrm{4}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}\:\:{and} \\ $$ $$\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\left(\mathrm{3}+\mathrm{2}{cosx}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}\pi}{\left(\mathrm{9}−\mathrm{4}\right)\sqrt{\mathrm{9}−\mathrm{4}}}\:=\frac{\mathrm{3}\pi}{\mathrm{5}\sqrt{\mathrm{5}}}\:. \\ $$

Commented byaliesam last updated on 11/Jul/19

Commented byaliesam last updated on 11/Jul/19

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