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Question Number 32039 by abdo imad last updated on 18/Mar/18
a>−1 calculate ∫_0 ^(π/2) (dt/(1+a tan^2 t)) . 2) find ∫_0 ^(π/2) ((tan^2 t)/((1+atan^2 t)^2 )) dt 3) find the value of ∫_0 ^(π/2) ((tan^2 t)/((1+2tan^2 t)^2 ))dt.
$${a}>−\mathrm{1}\:\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{a}\:{tan}^{\mathrm{2}} {t}}\:. \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{atan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:{dt} \\ $$$$\left.\mathrm{3}\right)\:\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+\mathrm{2}{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}.\: \\ $$
Commented by abdo imad last updated on 22/Mar/18
let put f(a) = ∫_0 ^(π/2) (dt/(1+atan^2 t)) f(a) = ∫_0 ^(π/2) (dt/(1+a((sin^2 t)/(cos^2 t)))) = ∫_0 ^(π/2) ((cos^2 t)/(cos^2 t +asin^2 t))dt f(a)= ∫_0 ^(π/2) (((1+cos(2t))/2)/(((1+cos(2t)/2) +a((1−cos(2t))/2)))dt = ∫_0 ^(π/2) ((1+cos(2t))/(1+cos(2t) +a(1−cos(2t)))) dt = ∫_0 ^(π/2) ((1+cos(2t))/(1+a +(1−a)cos(2t))) thech. 2t =u give f(a) = (1/2)∫_0 ^π ((1+cos(u))/(1+a +(1−a)cos(u)))du ch. tan((u/2))=x give f(a) =(1/2) ∫_0 ^∞ ((1+((1−x^2 )/(1+x^2 )))/(1+a +(1−a)((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) f(a) = ∫_0 ^∞ (2/((1+x^2 )((1+a)(1+x^2 ) +(1−a)(1−x^2 )))dx = ∫_0 ^∞ ((2dx)/((1+x^2 )( 1+a +(1+a)x^2 +(1−a) +(a−1)x^2 ))) = ∫_0 ^∞ ((2dx)/((1+x^2 )( 2 +2ax^2 ))) =∫_0 ^∞ (dx/((1+x^2 )(1+ax^2 ))) =(1/(1−a)) ∫_0 ^∞ ( (1/(1+x^2 )) − (a/(1+ax^2 )))dx f(a) = (π/(2(1−a))) − (a/(1−a)) ∫_0 ^∞ (dx/(1+ax^(2 ) )) but we have by ch. (√a) x = α (we take a>0) ∫_0 ^∞ (dx/(1+ax^2 )) = ∫_0 ^∞ (1/(1+α^2 )) (dα/( (√a))) =(π/(2(√a))) ⇒ f(a) = (π/(2(1−a))) −(a/(1−a)) (π/(2(√a))) f(a) = (π/(2(1−a)))( 1− (√a) ) ⇒ f(a)= (π/(2(1+(√a)))) .
$${let}\:{put}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{atan}^{\mathrm{2}} {t}} \\ $$$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{a}\frac{{sin}^{\mathrm{2}} {t}}{{cos}^{\mathrm{2}} {t}}}\:=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{{cos}^{\mathrm{2}} {t}}{{cos}^{\mathrm{2}} {t}\:+{asin}^{\mathrm{2}} {t}}{dt} \\ $$$${f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}{\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right.}{\mathrm{2}}\:+{a}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\:+{a}\left(\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)\right)}\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right){cos}\left(\mathrm{2}{t}\right)}\:\:{thech}.\:\mathrm{2}{t}\:={u}\:{give} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{\mathrm{1}+{cos}\left({u}\right)}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right){cos}\left({u}\right)}{du}\:\:{ch}.\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)={x}\:{give} \\ $$$${f}\left({a}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}+\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right)\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right.}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{2}{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\:\mathrm{1}+{a}\:+\left(\mathrm{1}+{a}\right){x}^{\mathrm{2}} \:+\left(\mathrm{1}−{a}\right)\:+\left({a}−\mathrm{1}\right){x}^{\mathrm{2}} \right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\frac{\mathrm{2}{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\:\mathrm{2}\:\:+\mathrm{2}{ax}^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{a}}\:\int_{\mathrm{0}} ^{\infty} \:\left(\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:−\:\:\frac{{a}}{\mathrm{1}+{ax}^{\mathrm{2}} }\right){dx} \\ $$$${f}\left({a}\right)\:=\:\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{a}\right)}\:−\:\frac{{a}}{\mathrm{1}−{a}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\mathrm{1}+{ax}^{\mathrm{2}\:} }\:{but}\:{we}\:{have} \\ $$$${by}\:{ch}.\:\sqrt{{a}}\:{x}\:=\:\alpha\:\:\left({we}\:{take}\:{a}>\mathrm{0}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{ax}^{\mathrm{2}} }\:=\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\frac{{d}\alpha}{\:\sqrt{{a}}}\:\:=\frac{\pi}{\mathrm{2}\sqrt{{a}}}\:\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{a}\right)}\:−\frac{{a}}{\mathrm{1}−{a}}\:\frac{\pi}{\mathrm{2}\sqrt{{a}}} \\ $$$${f}\left({a}\right)\:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{a}\right)}\left(\:\mathrm{1}−\:\sqrt{{a}}\:\right)\:\Rightarrow\:{f}\left({a}\right)=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\sqrt{{a}}\right)}\:. \\ $$
Commented by prof Abdo imad last updated on 22/Mar/18
2) we have proved that ∫_0 ^(π/2) (dt/(1+atan^2 t)) =f(a) = (π/(2(1+(√a)))) let derivate f^′ (a) = −∫_0 ^(π/2) ((tan^2 t)/((1+a tan^2 t)^2 ))dt ⇒ ∫_0 ^(π/2) ((tan^2 t)/((1+a tan^2 t)^2 ))dt =−f^′ (a) but f^′ (a) = −(π/2)(((1+(√a))^′ )/((1+(√a))^2 )) ) =−(π/2) (1/(2(√a)(1+(√a))^2 )) ⇒ ∫_0 ^(π/2) ((tan^2 t)/((1+atan^2 t)))dt = (π/(4(√a)(1+(√a))^2 )) 3) a=2 ⇒ ∫_0 ^(π/2) ((tan^2 t)/((1+2tan^2 t))) dt = (π/(4(√(2())1+(√2))^2 )) .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{proved}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{1}+{atan}^{\mathrm{2}} {t}}\:={f}\left({a}\right)\:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\sqrt{{a}}\right)}\:{let}\:{derivate} \\ $$$${f}^{'} \left({a}\right)\:=\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{a}\:{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{a}\:{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}\:=−{f}^{'} \left({a}\right)\:\:{but}\: \\ $$$${f}^{'} \left({a}\right)\:=\:−\frac{\pi}{\mathrm{2}}\left(\frac{\left.\mathrm{1}+\sqrt{{a}}\right)^{'} }{\left(\mathrm{1}+\sqrt{{a}}\right)^{\mathrm{2}} }\:\right)\:=−\frac{\pi}{\mathrm{2}}\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}\left(\mathrm{1}+\sqrt{{a}}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{atan}^{\mathrm{2}} {t}\right)}{dt}\:=\:\:\frac{\pi}{\mathrm{4}\sqrt{{a}}\left(\mathrm{1}+\sqrt{{a}}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{a}=\mathrm{2}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+\mathrm{2}{tan}^{\mathrm{2}} {t}\right)}\:{dt}\:=\:\:\:\frac{\pi}{\left.\mathrm{4}\sqrt{\mathrm{2}\left(\right.}\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:. \\ $$
Commented by abdo imad last updated on 22/Mar/18
a>0.
$${a}>\mathrm{0}. \\ $$

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