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

$$$$$$\phi \left(t\right)={\int }_0^{\frac{\pi }{2}}{(sin\left(x\right)+tcos\left(x\right))}^{2}dx$$$$findthevalueoftheextermum$$$$of\phi \left(t\right).$$

Answered by aleks041103 last updated on 04/Feb/22

$$sin\left(x\right)+tcos\left(x\right)=\sqrt{1+t^2}sin(x+arctan\left(t\right))$$$$\Rightarrow \phi \left(t\right)={\int }_0^{\pi /2}\frac{{sin}^2(x+a)}{1+t^2}dx=$$$$=(1+t^2){\int }_a^{a+\pi /2}{sin}^{2}xdx,a={tan}^{-1}t$$$$\int {sin}^{2}xdx=\int \frac{1-cos\left(2x\right)}{2}dx=$$$$=\frac{x}{2}-\frac{1}{4}sin\left(2x\right)$$$$\Rightarrow {\int }_a^{a+\pi /2}{sin}^{2}xdx=\frac{\pi }{4}-\frac{1}{4}(sin(\pi +2a)-sin\left(2a\right))=$$$$=\frac{\pi }{4}+sin\left(a\right)cos\left(a\right)=$$$$=\frac{\pi }{4}+tan\left(a\right){cos}^2\left(a\right)=$$$$=\frac{\pi }{4}+\frac{t}{1+{tan}^2\left(a\right)}=\frac{\pi }{4}+\frac{t}{1+t^2}$$$$\Rightarrow \phi \left(t\right)=\frac{\pi }{4}(1+t^2)+t$$$$$$

Commented by aleks041103 last updated on 04/Feb/22

$$\phi \left(t\right)=\frac{\pi }{4}{t}^2+t+\frac{\pi }{4}$$$${\phi }^{\prime }=\frac{\pi }{2}t+1=0$$$$\Rightarrow t=-\frac{2}{\pi }$$$$\Rightarrow {\phi }_{extr}=\frac{\pi }{4}\frac{4}{{\pi }^2}-\frac{2}{\pi }+\frac{\pi }{4}=$$$$=-\frac{1}{\pi }+\frac{\pi }{4}=\frac{{\pi }^2-4}{4\pi }={\phi }_{extr.}$$

Answered by mahdipoor last updated on 05/Feb/22

$${\phi }^{{}^{\prime }}\left(t\right)=\underset{k\to t}{\lim }\frac{\phi \left(k\right)-\phi \left(t\right)}{x-t}=$$$$\underset{k\to t}{\lim }\frac{{\int }_0^{\pi /2}(k^2-t^2){cos}^{2}x+2(k-t)sinx.cosx}{k-t}=$$$$\underset{k\to t}{\lim }{\int }_0^{\pi /2}(k+t){cos}^{2}x+2sinx.cosx=$$$$2t{\int }_0^{\pi /2}{cos}^{2}x+{\int }_0^{\pi /2}2sinx.cosx=\frac{\pi }{2}t+1$$$$\Rightarrow {\phi }^{{}^{\prime }}\left(t\right)=0\Rightarrow t=-\frac{2}{\pi }$$$$\Rightarrow \phi \left(t\right)=\int {\phi }^{{}^{\prime }}\left(t\right).dt=\frac{\pi }{4}{t}^2+t+c$$$$\Rightarrow \phi \left(0\right)={\int }_0^{\pi /2}{\left(sinx\right)}^2=\frac{\pi }{4}\Rightarrow c=\frac{\pi }{4}$$$$min\phi =\phi (-\frac{2}{\pi })=\frac{{\pi }^2-4}{4\pi }$$$$$$

Commented by aleks041103 last updated on 04/Feb/22

$$Onthesecondline:$$$$\frac{{\int }_0^{\pi /2}[(k^2-t^2){cos}^{2}x+2(k-t)sinx.cosx]dx}{k-t}$$$$since$$$$\phi \left(t\right)={\int }_0^{\pi /2}{(sin\left(x\right)+tcos\left(x\right))}^{2}dx$$

Commented by mahdipoor last updated on 04/Feb/22

$$thanksforyourhint$$

Commented by mahdipoor last updated on 05/Feb/22

$$bozorgvariaziz$$

Commented by mnjuly1970 last updated on 04/Feb/22

$$aliboodjenabemahdipoor.$$$$mamnoon.$$

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