t-0-pi-2-sin-x-t-cos-x-2-dx-find-the-value-of-the-extermum-of-t-

Question Number 165581 by mnjuly1970 last updated on 04/Feb/22

ϕ(t)=∫_0 ^( (π/2)) ( sin(x)+t cos(x))^( 2) dx find the value of the extermum of ϕ (t).

$$ \\ $$$$\varphi\left({t}\right)=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left(\:{sin}\left({x}\right)+{t}\:{cos}\left({x}\right)\right)^{\:\mathrm{2}} {dx} \\ $$$${find}\:\:{the}\:\:{value}\:{of}\:{the}\:{extermum} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{of}\:\:\:\varphi\:\left({t}\right). \\ $$

Answered by aleks041103 last updated on 04/Feb/22

sin(x)+tcos(x)=(√(1+t^2 ))sin(x+arctan(t)) ⇒ϕ(t)=∫_0 ^(π/2) ((sin^2 (x+a))/(1+t^2 ))dx= =(1+t^2 )∫_a ^(a+π/2) sin^2 x dx , a=tan^(−1) t ∫sin^2 x dx= ∫((1−cos(2x))/2)dx= =(x/2)−(1/4)sin(2x) ⇒∫_a ^(a+π/2) sin^2 x dx=(π/4)−(1/4)(sin(π+2a)−sin(2a))= =(π/4)+sin(a)cos(a)= =(π/4)+tan(a)cos^2 (a)= =(π/4)+(t/(1+tan^2 (a)))=(π/4)+(t/(1+t^2 )) ⇒ϕ(t)=(π/4)(1+t^2 )+t

$${sin}\left({x}\right)+{tcos}\left({x}\right)=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{sin}\left({x}+{arctan}\left({t}\right)\right) \\ $$$$\Rightarrow\varphi\left({t}\right)=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{sin}^{\mathrm{2}} \left({x}+{a}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dx}= \\ $$$$=\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\int_{{a}} ^{{a}+\pi/\mathrm{2}} {sin}^{\mathrm{2}} {x}\:{dx}\:,\:{a}={tan}^{−\mathrm{1}} {t} \\ $$$$\int{sin}^{\mathrm{2}} {x}\:{dx}=\:\int\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{dx}= \\ $$$$=\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{x}\right) \\ $$$$\Rightarrow\int_{{a}} ^{{a}+\pi/\mathrm{2}} {sin}^{\mathrm{2}} {x}\:{dx}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\left({sin}\left(\pi+\mathrm{2}{a}\right)−{sin}\left(\mathrm{2}{a}\right)\right)= \\ $$$$=\frac{\pi}{\mathrm{4}}+{sin}\left({a}\right){cos}\left({a}\right)= \\ $$$$=\frac{\pi}{\mathrm{4}}+{tan}\left({a}\right){cos}^{\mathrm{2}} \left({a}\right)= \\ $$$$=\frac{\pi}{\mathrm{4}}+\frac{{t}}{\mathrm{1}+{tan}^{\mathrm{2}} \left({a}\right)}=\frac{\pi}{\mathrm{4}}+\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\varphi\left({t}\right)=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{t} \\ $$$$ \\ $$

Commented by aleks041103 last updated on 04/Feb/22

ϕ(t)=(π/4)t^2 +t+(π/4) ϕ′=(π/2)t+1=0 ⇒t=−(2/π) ⇒ϕ_(extr) =(π/4) (4/π^2 )−(2/π)+(π/4)= =−(1/π)+(π/4)=((π^2 −4)/(4π))=ϕ_(extr.)

$$\varphi\left({t}\right)=\frac{\pi}{\mathrm{4}}{t}^{\mathrm{2}} +{t}+\frac{\pi}{\mathrm{4}} \\ $$$$\varphi'=\frac{\pi}{\mathrm{2}}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=−\frac{\mathrm{2}}{\pi} \\ $$$$\Rightarrow\varphi_{{extr}} =\frac{\pi}{\mathrm{4}}\:\frac{\mathrm{4}}{\pi^{\mathrm{2}} }−\frac{\mathrm{2}}{\pi}+\frac{\pi}{\mathrm{4}}= \\ $$$$=−\frac{\mathrm{1}}{\pi}+\frac{\pi}{\mathrm{4}}=\frac{\pi^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}\pi}=\varphi_{{extr}.} \\ $$

Answered by mahdipoor last updated on 05/Feb/22

ϕ^′ (t)=lim_(k→t) ((ϕ(k)−ϕ(t))/(x−t))= lim_(k→t) ((∫_0 ^( π/2) (k^2 −t^2 )cos^2 x+2(k−t)sinx.cosx)/(k−t))= lim_(k→t) ∫_0 ^( π/2) (k+t)cos^2 x+2sinx.cosx = 2t∫_0 ^( π/2) cos^2 x +∫_0 ^( π/2) 2sinx.cosx=(π/2)t+1 ⇒ ϕ^′ (t)=0 ⇒ t=−(2/π) ⇒ϕ(t)=∫ϕ^′ (t).dt=(π/4)t^2 +t+c ⇒ϕ(0)=∫_0 ^( π/2) (sinx)^2 =(π/4) ⇒c=(π/4) min ϕ = ϕ(−(2/π))=((π^2 −4)/(4π))

$$\varphi^{'} \left({t}\right)=\underset{{k}\rightarrow{t}} {\mathrm{lim}}\frac{\varphi\left({k}\right)−\varphi\left({t}\right)}{{x}−{t}}= \\ $$$$\underset{{k}\rightarrow{t}} {\mathrm{lim}}\frac{\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left({k}^{\mathrm{2}} −{t}^{\mathrm{2}} \right){cos}^{\mathrm{2}} {x}+\mathrm{2}\left({k}−{t}\right){sinx}.{cosx}}{{k}−{t}}= \\ $$$$\underset{{k}\rightarrow{t}} {\mathrm{lim}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left({k}+{t}\right){cos}^{\mathrm{2}} {x}+\mathrm{2}{sinx}.{cosx}\:= \\ $$$$\mathrm{2}{t}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {cos}^{\mathrm{2}} {x}\:+\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{2}{sinx}.{cosx}=\frac{\pi}{\mathrm{2}}{t}+\mathrm{1} \\ $$$$\Rightarrow\:\varphi^{'} \left({t}\right)=\mathrm{0}\:\Rightarrow\:{t}=−\frac{\mathrm{2}}{\pi} \\ $$$$\Rightarrow\varphi\left({t}\right)=\int\varphi^{'} \left({t}\right).{dt}=\frac{\pi}{\mathrm{4}}{t}^{\mathrm{2}} +{t}+{c} \\ $$$$\Rightarrow\varphi\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left({sinx}\right)^{\mathrm{2}} =\frac{\pi}{\mathrm{4}}\:\Rightarrow{c}=\frac{\pi}{\mathrm{4}} \\ $$$${min}\:\varphi\:=\:\varphi\left(−\frac{\mathrm{2}}{\pi}\right)=\frac{\pi^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}\pi} \\ $$$$ \\ $$

Commented by aleks041103 last updated on 04/Feb/22

On the second line: ((∫_0 ^( π/2) [(k^2 −t^2 )cos^2 x+2(k−t)sinx.cosx]dx)/(k−t)) since ϕ(t)=∫_0 ^(π/2) (sin(x)+t cos(x))^2 dx

$${On}\:{the}\:{second}\:{line}: \\ $$$$\frac{\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left[\left({k}^{\mathrm{2}} −{t}^{\mathrm{2}} \right){cos}^{\mathrm{2}} {x}+\mathrm{2}\left({k}−{t}\right){sinx}.{cosx}\right]{dx}}{{k}−{t}} \\ $$$${since} \\ $$$$\varphi\left({t}\right)=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({sin}\left({x}\right)+{t}\:{cos}\left({x}\right)\right)^{\mathrm{2}} {dx} \\ $$

Commented by mahdipoor last updated on 04/Feb/22