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Question Number 41848 by maxmathsup by imad last updated on 13/Aug/18
let f(a) = ∫_0 ^(π/2) (dx/(1+asinx)) with a∈R 1) find a simple form of f(a) 2) calculate ∫_0 ^(π/2) (dx/(1+sinx)) and ∫_0 ^(π/2) (dx/(1+2sinx)) 3) find the value of ∫_0 ^(π/2) ((cosx)/((1+asinx)^2 ))dx 4) find the value of ∫_0 ^(π/2) ((cosx)/((1+sinx)^2 ))dx and ∫_0 ^(π/2) ((cosx)/((1+2sinx)^2 ))dx
$${let}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\mathrm{1}+{asinx}}\:\:\:{with}\:{a}\in{R} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}+{sinx}}\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+\mathrm{2}{sinx}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cosx}}{\left(\mathrm{1}+{asinx}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{cosx}}{\left(\mathrm{1}+{sinx}\right)^{\mathrm{2}} }{dx}\:{and}\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cosx}}{\left(\mathrm{1}+\mathrm{2}{sinx}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 14/Aug/18
1) changement tan((x/2))=t give f(a) = ∫_0 ^1 (1/(1+a((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^1 ((2dt)/(1+t^2 +2at)) =∫_0 ^1 ((2dt)/(t^2 +2at +1)) roots of t^2 +2at +1 Δ^′ =a^2 −1 case 1 a^2 >1 ⇒ ∣a∣>1 ⇒ t_1 =−a +(√(a^2 −1)) qnd t_2 =−a−(√(a^2 −1)) and F(t) =(2/(t^2 +2at +1)) =(2/((t−t_1 )(t−t_2 ))) =(α/(t−t_1 )) +(β/(t−t_2 )) α = (2/(t_1 −t_2 )) = (2/(2(√(a^2 −1)))) = (1/( (√(a^2 −1)))) and β =(2/(t_2 −t_1 )) =−(1/( (√(a^2 −1)))) ⇒ F(t) =(1/( (√(a^2 −1)))){ (1/(t−t_1 )) −(1/(t−t_2 ))} ⇒f(a) =∫_0 ^1 (1/( (√(a^2 −1)))){ (1/(t−t_1 ))−(1/(t−t_2 ))}dt = (1/( (√(a^2 −1)))) [ln∣ ((t−t_1 )/(t−t_2 ))∣]_0 ^1 = (1/( (√(a^2 −1)))) { ln∣ ((1+a−(√(a^2 −1)))/(1+a+(√(a^2 −1))))∣−ln∣ ((a−(√(a^2 −1)))/(a+(√(a^2 −1))))∣} case 2 a^2 <1 ⇒ −1<a<1 ⇒ no real roots ⇒ f(a) =∫_0 ^1 ((2dt)/(t^2 +2at +a^2 +1−a^2 )) = ∫_0 ^1 ((2dt)/((t+a)^2 +1−a^2 )) changement t+a =(√(1−a^2 ))u give f(a) = ∫_(a/( (√(1−a^2 )))) ^((1+a)/( (√(1−a^2 )))) ((2 (√(1−a^2 ))du)/((1−a^2 )(1+u^2 ))) = (2/( (√(1−a^2 )))) { arctan(((1+a)/( (√(1−a^2 )))))−arctan((a/( (√(1−a^2 )))))}
$$\left.\mathrm{1}\right)\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{a}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{at}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{at}\:+\mathrm{1}} \\ $$$${roots}\:{of}\:{t}^{\mathrm{2}} \:+\mathrm{2}{at}\:+\mathrm{1} \\ $$$$\Delta^{'} \:={a}^{\mathrm{2}} −\mathrm{1}\:\:\: \\ $$$${case}\:\mathrm{1}\:\:\:{a}^{\mathrm{2}} >\mathrm{1}\:\Rightarrow\:\mid{a}\mid>\mathrm{1}\:\:\Rightarrow\:{t}_{\mathrm{1}} =−{a}\:+\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}}\:\:{qnd}\:{t}_{\mathrm{2}} =−{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\:\:{and} \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{2}}{{t}^{\mathrm{2}} \:+\mathrm{2}{at}\:+\mathrm{1}}\:=\frac{\mathrm{2}}{\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}\:=\frac{\alpha}{{t}−{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}−{t}_{\mathrm{2}} } \\ $$$$ \\ $$$$\alpha\:=\:\frac{\mathrm{2}}{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }\:=\:\frac{\mathrm{2}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}}}\:\:{and}\:\beta\:=\frac{\mathrm{2}}{{t}_{\mathrm{2}} −{t}_{\mathrm{1}} }\:=−\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\left\{\:\:\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\}\:\Rightarrow{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\left\{\:\:\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\}{dt} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:\:\left[{ln}\mid\:\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}}}\:\left\{\:{ln}\mid\:\frac{\mathrm{1}+{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{1}+{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\mid−{ln}\mid\:\frac{{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\mid\right\} \\ $$$${case}\:\mathrm{2}\:\:{a}^{\mathrm{2}} <\mathrm{1}\:\Rightarrow\:−\mathrm{1}<{a}<\mathrm{1}\:\:\Rightarrow\:{no}\:{real}\:{roots}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{at}\:+{a}^{\mathrm{2}} \:+\mathrm{1}−{a}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{dt}}{\left({t}+{a}\right)^{\mathrm{2}} \:+\mathrm{1}−{a}^{\mathrm{2}} }\:\:{changement} \\ $$$${t}+{a}\:=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }{u}\:{give}\:{f}\left({a}\right)\:=\:\int_{\frac{{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}} ^{\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}} \:\:\:\:\:\:\frac{\mathrm{2}\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }{du}}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:\:\left\{\:{arctan}\left(\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)\right\} \\ $$
Commented by maxmathsup by imad last updated on 14/Aug/18
2) ∫_0 ^(π/2) (dx/(1+2sinx)) =f(2) = (1/( (√3))){ ln∣ ((3−(√3))/(3+(√3)))∣ −ln∣ ((2−(√3))/(2+(√3)))∣} =(1/( (√3))){ ln(((3−(√3))/(3+(√3))))−ln(((2−(√3))/(2+(√3))))} let calculate ∫_0 ^(π/2) (dx/(1+sinx)) ∫_0 ^(π/2) (dx/(1+sinx)) =_(tan((x/2)) =u) ∫_0 ^1 (1/(1+((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^1 ((2du)/(1+u^2 +2t)) = ∫_0 ^1 ((2du)/((u+1)^2 )) =[((−2)/(1+u))]_0 ^1 =−2((1/2) −1)= 1 .
$$\left.\mathrm{2}\right)\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+\mathrm{2}{sinx}}\:={f}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{\:{ln}\mid\:\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{3}}}\mid\:−{ln}\mid\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{\:{ln}\left(\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{3}}}\right)−{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right)\right\}\:\:{let}\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}+{sinx}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+{sinx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={u}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{2}{t}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }\:=\left[\frac{−\mathrm{2}}{\mathrm{1}+{u}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\:−\mathrm{1}\right)=\:\mathrm{1}\:. \\ $$
Commented by maxmathsup by imad last updated on 15/Aug/18
3) we have f(a) = ∫_0 ^(π/2) (dx/(1+a sinx)) ⇒f^′ (a) =−∫_0 ^(π/2) ((sinx)/((1+asinx)^2 )) dx ⇒ ∫_0 ^(π/2) ((sinx)/((1+a sinx)^2 )) dx =−f^′ (a) if −1<a<1 f(a) = (2/( (√(1−a^2 )))){ arctan(((1+a)/( (√(1−a^2 )))))−arctan((a/( (√(1−a^2 ))))) ⇒ f^′ (a) = 2 (−((−a)/( (√(1−a^2 )))))×(1/((1−a^2 ))){ arctan(((1+a)/( (√(1−a^2 )))))−arctan((a/( (√(1−a^2 )))))} +(2/( (√(1−a^2 )))) { (((((1+a)/( (√(1−a^2 )))))^′ )/(1+(((1+a)/( (√(1−a^2 )))))^2 )) −((((a/( (√(1−a^2 )))))^′ )/(1+((a/( (√(1−a^2 )))))^2 ))} but (((1+a)/( (√(1−a^2 )))))^′ =(((√(1−a^2 )) −(1+a)(((−2a)/(2(√(1−a^2 ))))))/(1−a^2 )) = ((1−a^2 +a+a^2 )/((1−a^2 )(√(1−a^2 )))) =((1+a)/((1−a^2 )(√(1−a^2 )))) ((a/( (√(1−a^2 )))))^′ = (((√(1−a^2 )) −a (((−a)/( (√(1−a^2 ))))))/((1−a^2 ))) =((1−a^2 +a^2 )/((1−a^2 )(√(1−a^2 )))) =(1/((1−a^2 )(√(1−a^2 )))) ⇒ f^′ (a) = ((2a)/((1−a^2 )(√(1−a^2 )))){ arctan(((1+a)/( (√(1−a^2 ))))) −arctan((a/( (√(1−a^2 )))))} +(2/( (√(1−a^2 )))){ ((1+a)/((1−a^2 )(√(1−a^2 ))(1+(((1+a)^2 )/(1−a^2 ))))) − (1/((1−a^2 )(√(1−a^2 ))(1+(a^2 /(1−a^2 )))))} if ∣a∣>1 we follow the same way .
$$\left.\mathrm{3}\right)\:{we}\:{have}\:\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\mathrm{1}+{a}\:{sinx}}\:\Rightarrow{f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sinx}}{\left(\mathrm{1}+{asinx}\right)^{\mathrm{2}} }\:{dx} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sinx}}{\left(\mathrm{1}+{a}\:{sinx}\right)^{\mathrm{2}} }\:{dx}\:=−{f}^{'} \left({a}\right) \\ $$$${if}\:−\mathrm{1}<{a}<\mathrm{1}\:\:{f}\left({a}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\left\{\:{arctan}\left(\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)\:\Rightarrow\right. \\ $$$${f}^{'} \left({a}\right)\:=\:\mathrm{2}\:\left(−\frac{−{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)×\frac{\mathrm{1}}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\left\{\:{arctan}\left(\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)\right\} \\ $$$$+\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:\left\{\:\:\:\:\:\frac{\left(\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)^{'} }{\mathrm{1}+\left(\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)^{\mathrm{2}} }\:−\frac{\left(\frac{{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)^{'} }{\mathrm{1}+\left(\frac{{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)^{\mathrm{2}} }\right\}\:\:{but} \\ $$$$\left(\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)^{'} \:\:=\frac{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\:−\left(\mathrm{1}+{a}\right)\left(\frac{−\mathrm{2}{a}}{\mathrm{2}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)}{\mathrm{1}−{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}−{a}^{\mathrm{2}} \:+{a}+{a}^{\mathrm{2}} }{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:=\frac{\mathrm{1}+{a}}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }} \\ $$$$\left(\frac{{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)^{'} \:=\:\frac{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\:\:−{a}\:\left(\frac{−{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}−{a}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:=\:\frac{\mathrm{2}{a}}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\left\{\:{arctan}\left(\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)\:−{arctan}\left(\frac{{a}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\right)\right\} \\ $$$$+\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\left\{\:\:\:\:\:\frac{\mathrm{1}+{a}}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\left(\mathrm{1}+{a}\right)^{\mathrm{2}} }{\mathrm{1}−{a}^{\mathrm{2}} }\right)}\:−\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\mathrm{1}−{a}^{\mathrm{2}} }\right)}\right\} \\ $$$${if}\:\mid{a}\mid>\mathrm{1}\:{we}\:{follow}\:{the}\:{same}\:{way}\:. \\ $$$$ \\ $$

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