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Question Number 163288 by HongKing last updated on 05/Jan/22

Answered by Ar Brandon last updated on 05/Jan/22

I=∫_0 ^(π/2) (dx/(1+a^2 tan^2 x))=∫_0 ^(π/2) ((sec^2 x)/(sec^2 x+a^2 tan^2 xsec^2 x))dx =∫_0 ^∞ (dt/(1+t^2 +a^2 t^2 (1+t^2 )))=∫_0 ^∞ (dt/((t^2 +1)(a^2 t^2 +1))) =∫_0 ^∞ ((1/(1−a^2 ))∙(1/(t^2 +1))+(a^2 /(a^2 −1))∙(1/(a^2 t^2 +1)))dt =(1/(1−a^2 ))∙(π/2)+(a^2 /(a^2 −1))∙(1/a)∙(π/2)=((1−a)/(1−a^2 ))∙(π/2)=(π/(2(1+a)))

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{a}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} {x}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{2}} {x}}{\mathrm{sec}^{\mathrm{2}} {x}+{a}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} {x}\mathrm{sec}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} +{a}^{\mathrm{2}} {t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} −\mathrm{1}}\centerdot\frac{\mathrm{1}}{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}\right){dt} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\centerdot\frac{\pi}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} −\mathrm{1}}\centerdot\frac{\mathrm{1}}{{a}}\centerdot\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}−{a}}{\mathrm{1}−{a}^{\mathrm{2}} }\centerdot\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}\right)} \\ $$

Commented by Ar Brandon last updated on 05/Jan/22

(1/((t^2 +1)(a^2 t^2 +1)))=((pt+q)/(t^2 +1))+((rt+s)/(a^2 t^2 +1)) =(((pt+q)(a^2 t^2 +1)+(rt+s)(t^2 +1))/) lim_(t→i) =(pi+q)(1−a^2 )=1, p=0 , q=(1/(1−a^2 )) q+s=1⇒s=1−q=(a^2 /(a^2 −1)) p+r=0⇒r=0

$$\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{pt}+{q}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{rt}+{s}}{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\frac{\left({pt}+{q}\right)\left({a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}\right)+\left({rt}+{s}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{} \\ $$$$\underset{{t}\rightarrow{i}} {\mathrm{lim}}=\left({pi}+{q}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)=\mathrm{1},\:{p}=\mathrm{0}\:,\:{q}=\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$${q}+{s}=\mathrm{1}\Rightarrow{s}=\mathrm{1}−{q}=\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${p}+{r}=\mathrm{0}\Rightarrow{r}=\mathrm{0} \\ $$

Commented by HongKing last updated on 05/Jan/22

perfect solution my dear Sir thank you

$$\mathrm{perfect}\:\mathrm{solution}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by peter frank last updated on 06/Jan/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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