If ϕ ( x ) = 1 x ∫ π 4 x 4 2 sin t - 3 ϕ ' ( t ) d t , x > 0 then ϕ '…

Given,

$\varphi \left(x\right)=\frac{1}{\sqrt{x}}{\int }_{\pi /4}^x\left(4\sqrt{2}\sin t-3{\varphi }^{\text{'}}\left(t\right)\right)dt$

Differentiate both sides w.r.t. $x$ we get,

${\varphi }^{\text{'}}\left(x\right)=\frac{-1}{2x^{3/2}}{\int }_{\pi /4}^x\left(4\sqrt{2}\sin t-3{\varphi }^{\text{'}}\left(t\right)\right)dt+\frac{1}{\sqrt{x}}\left(4\sqrt{2}\sin x-3{\varphi }^{\text{'}}\left(x\right)\right)$

Put $x=\frac{\pi }{4}$ 

${\varphi }^{\text{'}}\left(\frac{\mathrm{\pi }}{4}\right)=\frac{-1}{2{\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}^{3/2}}\times 0+\sqrt{\frac{4}{\mathrm{\pi }}}\left(4\sqrt{2}\times \frac{1}{\sqrt{2}}-3{\varphi }^{\text{'}}\left(\frac{\mathrm{\pi }}{4}\right)\right)$

$\Rightarrow \sqrt{\frac{\mathrm{\pi }}{4}}{\varphi }^{\text{'}}\left(\frac{\mathrm{\pi }}{4}\right)+3{\varphi }^{\text{'}}\left(\frac{\mathrm{\pi }}{4}\right)=4$

$\Rightarrow {\varphi }^{\text{'}}\left(\frac{\pi }{4}\right)\left(3+\sqrt{\frac{\pi }{4}}\right)=4$

$\Rightarrow {\varphi }^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{8}{6+\sqrt{\pi }}$