The maximum value of $f\left(x\right)={\int }_0^1 t\sin (x+\pi t)dt$ is

We have,

$\begin{array}{l}f\left(x\right)={\int }_0^1 t\sin (x+\pi t)dt\\ \Rightarrow f\left(x\right)={\left[-\frac{t}{\pi }\cos (x+\pi t)\right]}_0^1+\frac{1}{\pi }{\int }_0^1 \cos (x+\pi t)dt\\ \Rightarrow f\left(x\right)=\frac{1}{\pi }\cos x+\frac{1}{{\pi }^2}[\sin (x+\pi t){]}_0^{1}dt\\ \Rightarrow f\left(x\right)=\frac{1}{\pi }\cos x-\frac{2}{{\pi }^2}\sin x\end{array}$

We know that

$\begin{array}{l}-\sqrt{a^2+b^2}\le a\cos t+b\sin t\le \sqrt{a^2+b^2}\text{ for all }t\\ \therefore -\sqrt{\frac{1}{{\pi }^2}+\frac{4}{{\pi }^4}}\le \frac{1}{\pi }\cos x\mid -\frac{2}{{\pi }^2}\sin x\le \sqrt{\frac{1}{{\pi }^2}+\frac{4}{{\pi }^4}}\text{ for all }x\\ \Rightarrow -\frac{\sqrt{{\pi }^2+4}}{{\pi }^2}\le f\left(x\right)\le \frac{\sqrt{{\pi }^2+4}}{{\pi }^2}\text{ for all }x\\ \Rightarrow f_{max}\left(x\right)=\frac{1}{{\pi }^2}\sqrt{{\pi }^2+4}\end{array}$