Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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 by aliesam last updated on 12/Jul/19

god bless you sir ..well done..

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

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

you are welcome.

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

Commented by aliesam last updated on 11/Jul/19

thank you sir

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

Commented by aliesam 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 by mathmax 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 by aliesam last updated on 11/Jul/19

Commented by aliesam last updated on 11/Jul/19

Terms of Service

Privacy Policy

Contact: info@tinkutara.com