Question Number 30775 by abdo imad last updated on 25/Feb/18
![letα ∈]0,π[ calculate ∫_0 ^(π/2) (dx/(2(cosα +chx))) .](Q30775.png)
$$\left.{let}\alpha\:\in\right]\mathrm{0},\pi\left[\:\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{2}\left({cos}\alpha\:+{chx}\right)}\:.\right. \\ $$
Commented by prof Abdo imad last updated on 25/Feb/18
![we have I = ∫_0 ^(π/2) (dx/(2(cosα +((e^x + e^(−x) )/2)))) = ∫_0 ^(π/2) (dx/(2cosα +e^x +e^(−x) )) the ch. e^x =t I = ∫_1 ^e^(π/2) (1/(2cosα +t +(1/t)))(dt/t) = ∫_1 ^e^(π/2) (dt/(2t cosα +t^2 +1))=∫_1 ^e^(π/2) (dt/(t^2 +2tcosα +1)) = ∫_0 ^e^(π/2) (dt/((t +cosα)^2 +sin^2 α)) let use the ch. t+cosα =u sinα ⇒u =(1/(sinα))(t +cosα) I= ∫_(cotan(α)) ^((e^(π/2) +cosα)/(sinα)) ((sinα du)/(sin^2 α(1+u^2 ))) = (1/(sinα)) [arctanu]_(cotanα) ^((e^(π/2) +cosα)/(sinα)) =(1/(sinα))( arctan( ((e^(π/2) +cosα)/(sinα))) −arctan(cotanα)).](Q30801.png)
$${we}\:{have}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{2}\left({cos}\alpha\:+\frac{{e}^{{x}} \:+\:{e}^{−{x}} }{\mathrm{2}}\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{{dx}}{\mathrm{2}{cos}\alpha\:+{e}^{{x}} \:+{e}^{−{x}} }\:\:{the}\:{ch}.\:{e}^{{x}} ={t} \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{{e}^{\frac{\pi}{\mathrm{2}}} } \:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{cos}\alpha\:+{t}\:+\frac{\mathrm{1}}{{t}}}\frac{{dt}}{{t}} \\ $$$$=\:\int_{\mathrm{1}} ^{{e}^{\frac{\pi}{\mathrm{2}}} } \:\:\:\:\:\:\:\frac{{dt}}{\mathrm{2}{t}\:{cos}\alpha\:+{t}^{\mathrm{2}} \:+\mathrm{1}}=\int_{\mathrm{1}} ^{{e}^{\frac{\pi}{\mathrm{2}}} } \:\:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{tcos}\alpha\:+\mathrm{1}} \\ $$$$=\:\int_{\mathrm{0}} ^{{e}^{\frac{\pi}{\mathrm{2}}} } \:\:\:\:\frac{{dt}}{\left({t}\:+{cos}\alpha\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \alpha}\:{let}\:{use}\:{the}\:{ch}. \\ $$$${t}+{cos}\alpha\:={u}\:{sin}\alpha\:\Rightarrow{u}\:=\frac{\mathrm{1}}{{sin}\alpha}\left({t}\:+{cos}\alpha\right) \\ $$$${I}=\:\int_{{cotan}\left(\alpha\right)} ^{\frac{{e}^{\frac{\pi}{\mathrm{2}}} \:+{cos}\alpha}{{sin}\alpha}} \:\:\:\:\:\:\:\:\:\frac{{sin}\alpha\:{du}}{{sin}^{\mathrm{2}} \alpha\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{1}}{{sin}\alpha}\:\left[{arctanu}\right]_{{cotan}\alpha} ^{\frac{{e}^{\frac{\pi}{\mathrm{2}}} \:+{cos}\alpha}{{sin}\alpha}} \\ $$$$=\frac{\mathrm{1}}{{sin}\alpha}\left(\:{arctan}\left(\:\:\frac{{e}^{\frac{\pi}{\mathrm{2}}} \:+{cos}\alpha}{{sin}\alpha}\right)\:−{arctan}\left({cotan}\alpha\right)\right). \\ $$$$ \\ $$
Answered by sma3l2996 last updated on 25/Feb/18
![A=∫_0 ^(π/2) (dx/(2(cosα+chx))) let t=th(x/2)⇒dt=(1/2)(1−th^2 (x/2))dx dx=(2/(1−t^2 ))dt ch(x)=2ch^2 (x/2)−1=(2/(1−th^2 (x/2)))−1=(2/(1−t^2 ))−1=((1+t^2 )/(1−t^2 )) A=∫_0 ^(th(π/4)) (1/(2(cosα+((1+t^2 )/(1−t^2 )))))×(((2dt)/(1−t^2 ))) =∫_0 ^(th(π/4)) (dt/((1−t^2 )cosα+1+t^2 ))=∫_0 ^(th(π/4)) (dt/(t^2 (1−cosα)+1+cosα)) A=∫_0 ^(th(π/4)) (dt/((1+cosα)((((1−cosα)/(1+cosα)))t^2 +1))) let′s put u=(((1−cosα)/(1+cosα)))^(1/2) t⇒dt=(((1+cosα)/(1−cosα)))du A=(1/((1−cosα)^(1/2) (1+cosα)^(1/2) ))∫_0 ^a (du/(u^2 +1)) A=(1/((1−cos^2 α)^(1/2) ))×[arctan(((1−cosα)/(sinα))t)]_0 ^(th(π/4)) A=(1/(sinα))arctan(((1−cosα)/(sinα))th(π/4))](Q30781.png)
$${A}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dx}}{\mathrm{2}\left({cos}\alpha+{chx}\right)} \\ $$$${let}\:\:{t}={th}\left({x}/\mathrm{2}\right)\Rightarrow{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{th}^{\mathrm{2}} \left({x}/\mathrm{2}\right)\right){dx} \\ $$$${dx}=\frac{\mathrm{2}}{\mathrm{1}−{t}^{\mathrm{2}} }{dt} \\ $$$${ch}\left({x}\right)=\mathrm{2}{ch}^{\mathrm{2}} \left({x}/\mathrm{2}\right)−\mathrm{1}=\frac{\mathrm{2}}{\mathrm{1}−{th}^{\mathrm{2}} \left({x}/\mathrm{2}\right)}−\mathrm{1}=\frac{\mathrm{2}}{\mathrm{1}−{t}^{\mathrm{2}} }−\mathrm{1}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$${A}=\int_{\mathrm{0}} ^{{th}\left(\pi/\mathrm{4}\right)} \frac{\mathrm{1}}{\mathrm{2}\left({cos}\alpha+\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\right)}×\left(\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{{th}\left(\pi/\mathrm{4}\right)} \frac{{dt}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right){cos}\alpha+\mathrm{1}+{t}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{{th}\left(\pi/\mathrm{4}\right)} \frac{{dt}}{{t}^{\mathrm{2}} \left(\mathrm{1}−{cos}\alpha\right)+\mathrm{1}+{cos}\alpha} \\ $$$${A}=\int_{\mathrm{0}} ^{{th}\left(\pi/\mathrm{4}\right)} \frac{{dt}}{\left(\mathrm{1}+{cos}\alpha\right)\left(\left(\frac{\mathrm{1}−{cos}\alpha}{\mathrm{1}+{cos}\alpha}\right){t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${let}'{s}\:{put}\:\:{u}=\left(\frac{\mathrm{1}−{cos}\alpha}{\mathrm{1}+{cos}\alpha}\right)^{\mathrm{1}/\mathrm{2}} {t}\Rightarrow{dt}=\left(\frac{\mathrm{1}+{cos}\alpha}{\mathrm{1}−{cos}\alpha}\right){du} \\ $$$${A}=\frac{\mathrm{1}}{\left(\mathrm{1}−{cos}\alpha\right)^{\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{cos}\alpha\right)^{\mathrm{1}/\mathrm{2}} }\int_{\mathrm{0}} ^{{a}} \frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$${A}=\frac{\mathrm{1}}{\left(\mathrm{1}−{cos}^{\mathrm{2}} \alpha\right)^{\mathrm{1}/\mathrm{2}} }×\left[{arctan}\left(\frac{\mathrm{1}−{cos}\alpha}{{sin}\alpha}{t}\right)\right]_{\mathrm{0}} ^{{th}\left(\pi/\mathrm{4}\right)} \\ $$$${A}=\frac{\mathrm{1}}{{sin}\alpha}{arctan}\left(\frac{\mathrm{1}−{cos}\alpha}{{sin}\alpha}{th}\left(\pi/\mathrm{4}\right)\right) \\ $$$$ \\ $$