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Question Number 43823 by abdo.msup.com last updated on 15/Sep/18

let ϕ(a,x) =∫_0 ^(π/2) (dt/(x+asin^2 t)) 2) find a exlicite form of ϕ(a,x) 3)determine ϕ(1,x)and ϕ(a,1) 4)find the vslue of ∫_0 ^(π/2) (dt/(2+3sin^2 t)) 5)find ∫_0 ^(π/2) (dt/((x+asin^2 t)^2 )) 6) find ∫_0 ^(π/2) ((sin^2 t)/((x+a sin^2 t)^2 ))dt .

$${let}\:\varphi\left({a},{x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{{x}+{asin}^{\mathrm{2}} {t}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{exlicite}\:{form}\:{of}\:\varphi\left({a},{x}\right) \\ $$$$\left.\mathrm{3}\right){determine}\:\varphi\left(\mathrm{1},{x}\right){and}\:\varphi\left({a},\mathrm{1}\right) \\ $$$$\left.\mathrm{4}\right){find}\:{the}\:{vslue}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{2}+\mathrm{3}{sin}^{\mathrm{2}} {t}} \\ $$$$\left.\mathrm{5}\right){find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\left({x}+{asin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{6}\right)\:{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sin}^{\mathrm{2}} {t}}{\left({x}+{a}\:{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}\:. \\ $$

Commented by maxmathsup by imad last updated on 17/Sep/18

1) we have ϕ(a,x) =∫_0 ^(π/2) (dt/(x+a ((1−cos(2t))/2))) = ∫_0 ^(π/2) ((2dt)/(2x +a −a cos(2t))) =_(2t =u) ∫_0 ^π (du/(2x +a−acosu)) =_(tan((u/2))=α) ∫_0 ^∞ (1/(2x+a−a((1−α^2 )/(1+α^2 )))) ((2dα)/(1+α^2 )) = ∫_0 ^∞ ((2 dα)/((2x+1)(1+α^2 )−a(1−α^2 ))) =∫_0 ^∞ ((2dα)/((2x+1) +(2x+1 +a)α^2 −a)) = ∫_0 ^∞ ((2dα)/((2x+1+a)α^2 + 2x+1−a)) =(2/((2x+1+a))) ∫_0 ^∞ (dα/(α^2 +((2x+1−a)/(2x+1+a)))) case 1 ((2x+1−a)/(2x+1+a))>0 changement α =(√((2x+1−a)/(2x+1+a)))z give ϕ(a,x) = (2/((2x+1+a))) ∫_0 ^∞ (1/(((2x+1−a)/(2x+1 +a))(1+z^2 ))) (√((2x+1−a)/(2x+1+a)))dz =(2/(2x+1+a)) ((2x+1+a)/(2x+1−a)) (√((2x+1−a)/(2x+1+a)))(π/2) =(π/(2x+1+a)) (√((2x+1+a)/(2x+1−a))) = (π/(√((2x+1+a)^2 −a^2 ))) case 2 ((2x+1−a)/(2x+1 +a))<0 ⇒ϕ(a,x) = (2/(2x+1+a)) ∫_0 ^∞ (dα/(α^2 −((a−(2x+1))/(a+2x+1)))) changement α =(√((a−(2x+1))/(a+2x+1)))z give ϕ(a,x) = (2/(2x+1+a)) ∫_0 ^∞ (1/(((a−(2x+1))/(a+2x+1))(z^2 −1))) (√((a−(2x+1))/(a+(2x+1)))) dz = (2/(2x+1+a)) ((a +2x+1)/(a−(2x+1))) (√((a−(2x+1))/(a+(2x+1))))∫_0 ^∞ (dz/(z^2 −1)) = (π/(√(a^2 −(2x+1)^2 ))) (1/2)∫_0 ^∞ {(1/(z−1)) −(1/(z+1))}dz =(π/(2(√(a^2 −(2x+1)^2 ))))[ln∣((z−1)/(z+1))∣]_0 ^(+∞) =0

$$\left.\mathrm{1}\right)\:{we}\:{have}\:\varphi\left({a},{x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{{x}+{a}\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}{x}\:+{a}\:−{a}\:{cos}\left(\mathrm{2}{t}\right)} \\ $$$$=_{\mathrm{2}{t}\:={u}} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{du}}{\mathrm{2}{x}\:+{a}−{acosu}}\:=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)=\alpha} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{x}+{a}−{a}\frac{\mathrm{1}−\alpha^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{2}} }}\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}\:{d}\alpha}{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)−{a}\left(\mathrm{1}−\alpha^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{d}\alpha}{\left(\mathrm{2}{x}+\mathrm{1}\right)\:+\left(\mathrm{2}{x}+\mathrm{1}\:+{a}\right)\alpha^{\mathrm{2}} −{a}} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{d}\alpha}{\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)\alpha^{\mathrm{2}} \:+\:\mathrm{2}{x}+\mathrm{1}−{a}}\:=\frac{\mathrm{2}}{\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{d}\alpha}{\alpha^{\mathrm{2}} \:+\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}} \\ $$$${case}\:\mathrm{1}\:\:\:\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}>\mathrm{0}\:{changement}\:\alpha\:=\sqrt{\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}}{z}\:{give} \\ $$$$\varphi\left({a},{x}\right)\:=\:\frac{\mathrm{2}}{\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}\:+{a}}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:\sqrt{\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}}{dz} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}+{a}}\:\frac{\mathrm{2}{x}+\mathrm{1}+{a}}{\mathrm{2}{x}+\mathrm{1}−{a}}\:\sqrt{\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}}\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\mathrm{2}{x}+\mathrm{1}+{a}}\:\sqrt{\frac{\mathrm{2}{x}+\mathrm{1}+{a}}{\mathrm{2}{x}+\mathrm{1}−{a}}} \\ $$$$=\:\frac{\pi}{\sqrt{\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$${case}\:\mathrm{2}\:\:\:\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}\:+{a}}<\mathrm{0}\:\Rightarrow\varphi\left({a},{x}\right)\:=\:\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}+{a}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{d}\alpha}{\alpha^{\mathrm{2}} −\frac{{a}−\left(\mathrm{2}{x}+\mathrm{1}\right)}{{a}+\mathrm{2}{x}+\mathrm{1}}} \\ $$$${changement}\:\:\alpha\:=\sqrt{\frac{{a}−\left(\mathrm{2}{x}+\mathrm{1}\right)}{{a}+\mathrm{2}{x}+\mathrm{1}}}{z}\:{give} \\ $$$$\varphi\left({a},{x}\right)\:=\:\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}+{a}}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\frac{{a}−\left(\mathrm{2}{x}+\mathrm{1}\right)}{{a}+\mathrm{2}{x}+\mathrm{1}}\left({z}^{\mathrm{2}} −\mathrm{1}\right)}\:\sqrt{\frac{{a}−\left(\mathrm{2}{x}+\mathrm{1}\right)}{{a}+\left(\mathrm{2}{x}+\mathrm{1}\right)}}\:{dz} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}+{a}}\:\frac{{a}\:+\mathrm{2}{x}+\mathrm{1}}{{a}−\left(\mathrm{2}{x}+\mathrm{1}\right)}\:\sqrt{\frac{{a}−\left(\mathrm{2}{x}+\mathrm{1}\right)}{{a}+\left(\mathrm{2}{x}+\mathrm{1}\right)}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dz}}{{z}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\:\frac{\pi}{\sqrt{{a}^{\mathrm{2}} −\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\left\{\frac{\mathrm{1}}{{z}−\mathrm{1}}\:−\frac{\mathrm{1}}{{z}+\mathrm{1}}\right\}{dz}\:=\frac{\pi}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} −\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }}\left[{ln}\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{0} \\ $$

Commented by maxmathsup by imad last updated on 17/Sep/18

2) for a =1 we get ((2x+1−a)/(2x+1+a)) =((2x)/(2x+2)) =(x/(x+1)) case 1 (x/(x+1))>0 ⇒ϕ(1,x) = (π/(√(4(x+1)^2 −1))) case 2 (x/(x+1))<0 ⇒ϕ(1,x) =0 also for x=1 we have ((2x+1−a)/(2x+1+a)) =((3−a)/(3+a)) case 1 ((3−a)/(3+a))>0 ⇒ ϕ(a,1) = (π/(√((3+a)^2 −a^2 ))) case 2 ((3−a)/(3+a))<0 ⇒ϕ(a,1)=0 .

$$\left.\mathrm{2}\right)\:{for}\:{a}\:=\mathrm{1}\:{we}\:{get}\:\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}\:=\frac{\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{2}}\:=\frac{{x}}{{x}+\mathrm{1}} \\ $$$${case}\:\mathrm{1}\:\:\frac{{x}}{{x}+\mathrm{1}}>\mathrm{0}\:\Rightarrow\varphi\left(\mathrm{1},{x}\right)\:=\:\frac{\pi}{\sqrt{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}} \\ $$$${case}\:\mathrm{2}\:\:\frac{{x}}{{x}+\mathrm{1}}<\mathrm{0}\:\Rightarrow\varphi\left(\mathrm{1},{x}\right)\:=\mathrm{0} \\ $$$${also}\:{for}\:{x}=\mathrm{1}\:\:{we}\:{have}\:\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}\:=\frac{\mathrm{3}−{a}}{\mathrm{3}+{a}} \\ $$$${case}\:\mathrm{1}\:\:\frac{\mathrm{3}−{a}}{\mathrm{3}+{a}}>\mathrm{0}\:\Rightarrow\:\varphi\left({a},\mathrm{1}\right)\:=\:\frac{\pi}{\sqrt{\left(\mathrm{3}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$${case}\:\mathrm{2}\:\:\frac{\mathrm{3}−{a}}{\mathrm{3}+{a}}<\mathrm{0}\:\Rightarrow\varphi\left({a},\mathrm{1}\right)=\mathrm{0}\:. \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 17/Sep/18

3) ∫_0 ^(π/2) (dt/(2+3sin^2 t)) =ϕ(3,2) wehave a=3 and x=2 we get ((2x+1−a)/(2x+1 +a)) =((5−3)/8) =(1/4)>0 ⇒ ϕ(3,2) = (π/(√((5+3)^2 −3^2 ))) =(π/(√(64−9))) =(π/(√(55)))

$$\left.\mathrm{3}\right)\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{2}+\mathrm{3}{sin}^{\mathrm{2}} {t}}\:=\varphi\left(\mathrm{3},\mathrm{2}\right)\:{wehave}\:{a}=\mathrm{3}\:{and}\:{x}=\mathrm{2} \\ $$$${we}\:{get}\:\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}\:+{a}}\:=\frac{\mathrm{5}−\mathrm{3}}{\mathrm{8}}\:=\frac{\mathrm{1}}{\mathrm{4}}>\mathrm{0}\:\Rightarrow\:\varphi\left(\mathrm{3},\mathrm{2}\right)\:=\:\frac{\pi}{\sqrt{\left(\mathrm{5}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }}\:=\frac{\pi}{\sqrt{\mathrm{64}−\mathrm{9}}}\:=\frac{\pi}{\sqrt{\mathrm{55}}} \\ $$

Commented by maxmathsup by imad last updated on 18/Sep/18

4) we have ϕ(a,x) = ∫_0 ^(π/2) (dt/(x+a sin^2 t)) ⇒ (∂ϕ/∂x)(a,x) =−∫_0 ^(π/2) (dt/((x+asin^2 t)^2 )) ⇒ ∫_0 ^(π/2) (dt/((x +asin^2 t)^2 )) =−(∂ϕ/∂x)(a,x) case 1 ((2x+1−a)/(2x+1+a))>0 ⇒ϕ(a,x) =π{(2x+1+a)^2 −a^2 }^(−(1/2)) ⇒ (∂ϕ/∂x)(a,x) =−(π/2)(4(2x+1+a)){(2x+1+a)^2 −a^2 }^(−(3/2)) = ((2π(2x+1+a))/(((2x+1+a)^2 −a^2 )(√((2x+1+a)^2 −a^2 )))) . case 2 ((zx+1−a)/(2x+1+a))<0 ⇒ϕ(a,x)=0 ⇒ (∂ϕ/∂x)(a,x)=0 .

$$\left.\mathrm{4}\right)\:{we}\:{have}\:\varphi\left({a},{x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{{x}+{a}\:{sin}^{\mathrm{2}} {t}}\:\Rightarrow\:\frac{\partial\varphi}{\partial{x}}\left({a},{x}\right)\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\left({x}+{asin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\left({x}\:+{asin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:=−\frac{\partial\varphi}{\partial{x}}\left({a},{x}\right) \\ $$$${case}\:\mathrm{1}\:\frac{\mathrm{2}{x}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}>\mathrm{0}\:\Rightarrow\varphi\left({a},{x}\right)\:=\pi\left\{\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right\}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow \\ $$$$\frac{\partial\varphi}{\partial{x}}\left({a},{x}\right)\:=−\frac{\pi}{\mathrm{2}}\left(\mathrm{4}\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)\right)\left\{\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right\}^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{2}\pi\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)}{\left(\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\sqrt{\left(\mathrm{2}{x}+\mathrm{1}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:. \\ $$$${case}\:\mathrm{2}\:\frac{{zx}+\mathrm{1}−{a}}{\mathrm{2}{x}+\mathrm{1}+{a}}<\mathrm{0}\:\Rightarrow\varphi\left({a},{x}\right)=\mathrm{0}\:\Rightarrow\:\frac{\partial\varphi}{\partial{x}}\left({a},{x}\right)=\mathrm{0}\:. \\ $$

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