The least value of the function $\phi(x)=\int\limits_{5 \pi / 4}^x(3 \sin t+4 \cos t) d t$ on the interval $[5 \pi / 4,4 \pi / 3]$, is

$-2 \sqrt{3}+\frac{3}{2}+\frac{1}{\sqrt{2}}$

We have,

$\phi(x) =\int\limits_{5 \pi / 4}^x(3 \sin t+4 \cos t) d t$

$\Rightarrow \phi'(x) =(3 \sin x+4 \cos x)$

Clearly, $\phi'(x)<0$ for all $x \in[5 \pi / 4,4 \pi / 3]$

$\Rightarrow \phi(x)$ is a decreasing function on $[5 \pi / 4,4 \pi / 3]$

$\Rightarrow \phi(4 \pi / 3)$ is the least value of $\phi(x)$ on $[5 \pi / 4,4 \pi / 3]$

Now,

$\phi\left(4 \frac{\pi}{3}\right) =\int\limits_{5 \pi / 4}^{4 \pi / 3}(3 \sin t+4 \cos t) d t$

$\Rightarrow \phi\left(4 \frac{\pi}{3}\right) =[-3 \cos t+4 \sin t]_{5 \pi / 4}^{4 \pi / 3}=\frac{3}{2}-2 \sqrt{3}+\frac{1}{\sqrt{2}}$