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Question Number 29446 by prof Abdo imad last updated on 08/Feb/18
let give a<1 find the value of f(a)= ∫_0 ^(π/2) (dx/(1−acos^2 x)).
$${let}\:{give}\:{a}<\mathrm{1}\:{find}\:{the}\:{value}\:{of} \\ $$$${f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}−{acos}^{\mathrm{2}} {x}}. \\ $$
Commented by prof Abdo imad last updated on 07/Mar/18
f(a)=∫_0 ^(π/2) (dx/(1−a((1 +cos(2x))/2)))=∫_0 ^(π/2) ((2dx)/(2−a −acos(2x))) = ∫_0 ^π (dt/(2−a −a cost)) and the ch.tan((t/2))=u give f(a)= ∫_0 ^∞ (1/(2−a−a ((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) =∫_0 ^∞ ((2du)/((2−a)(1+u^2 )−a(1−u^2 ))) =∫_0 ^∞ ((2du)/(2−a +(2−a)u^2 −a +au^2 )) =∫_0 ^∞ ((2du)/(2−2a +2u^2 )) =∫_0 ^∞ (du/(u^2 −a+1)) =∫_0 ^∞ (du/(u^2 +1−a)) the ch.u=(√(1−a)) x give f(a)=∫_0 ^∞ (((√(1−a)) dx)/((1−a)(1+x^2 )))=(1/( (√(1−a)))) (π/2) ⇒ f(a)= (π/(2(√(1−a)))) .
$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}−{a}\frac{\mathrm{1}\:+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{2}{dx}}{\mathrm{2}−{a}\:−{acos}\left(\mathrm{2}{x}\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}−{a}\:−{a}\:{cost}}\:{and}\:{the}\:{ch}.{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give} \\ $$$${f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}−{a}−{a}\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{2}−{a}\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)−{a}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}−{a}\:+\left(\mathrm{2}−{a}\right){u}^{\mathrm{2}} \:−{a}\:+{au}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}−\mathrm{2}{a}\:+\mathrm{2}{u}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:−{a}+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}−{a}}\:{the}\:{ch}.{u}=\sqrt{\mathrm{1}−{a}}\:{x}\:{give} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\sqrt{\mathrm{1}−{a}}\:{dx}}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{a}}}\:\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({a}\right)=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{1}−{a}}}\:\:. \\ $$
Answered by sma3l2996 last updated on 09/Feb/18
t=tanx⇒dx=(dt/(1+t^2 )) f(a)=∫_0 ^(+∞) (1/(1−a×(1/(1+t^2 ))))×(dt/(1+t^2 )) =∫_0 ^(+∞) (dt/(t^2 +1−a))=∫_0 ^(+∞) (dt/((1−a)(((t/( (√(1−a)))))^2 +1))) t=(√(1−a))u⇒dt=(√(1−a))du f(a)=((√(1−a))/(1−a))∫_0 ^∞ (du/(u^2 +1))=((√(1−a))/(1−a))[arctanu]_0 ^(+∞) f(a)=((√(1−a))/(1−a))×(π/2)
$${t}={tanx}\Rightarrow{dx}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{\mathrm{1}−{a}×\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }}×\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}−{a}}=\int_{\mathrm{0}} ^{+\infty} \frac{{dt}}{\left(\mathrm{1}−{a}\right)\left(\left(\frac{{t}}{\:\sqrt{\mathrm{1}−{a}}}\right)^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${t}=\sqrt{\mathrm{1}−{a}}{u}\Rightarrow{dt}=\sqrt{\mathrm{1}−{a}}{du} \\ $$$${f}\left({a}\right)=\frac{\sqrt{\mathrm{1}−{a}}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\sqrt{\mathrm{1}−{a}}}{\mathrm{1}−{a}}\left[{arctanu}\right]_{\mathrm{0}} ^{+\infty} \\ $$$${f}\left({a}\right)=\frac{\sqrt{\mathrm{1}−{a}}}{\mathrm{1}−{a}}×\frac{\pi}{\mathrm{2}} \\ $$

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