Mathematics - Differential Equation Question with Solution | TestHub

$f^{\text{'}}\left(x\right)={\tan }^{-1}\left(\sec x+\tan x\right)={\tan }^{-1}\left(\frac{1+\sin x}{\cos x}\right)={\tan }^{-1}\left(\frac{1-\cos \left(\frac{\pi }{2}+x\right)}{\sin \left(\frac{\pi }{2}+x\right)}\right)$

$\Rightarrow f^{\text{'}}\left(x\right)={\tan }^{-1}\left(\frac{2{\sin }^2\left(\frac{\pi }{4}+\frac{x}{2}\right)}{2\sin \left(\frac{\pi }{4}+\frac{x}{2}\right)\cos \left(\frac{\pi }{4}+\frac{x}{2}\right)}\right)$

$\Rightarrow \left(f^{\text{'}}\left(x\right)\right)dx=\left(\frac{\pi }{4}+\frac{x}{2}\right)dx$

$\Rightarrow f\left(x\right)=\frac{\pi }{4}x+\frac{x^2}{4}+c$

$\because f\left(0\right)=0\Rightarrow c=0$

So, $f\left(1\right)=\frac{\pi +1}{4}$