Given function

$f(x)=\int_0^x \sqrt{t}\sin t\,dt, \qquad x\in \left(0,\frac{5\pi}{2}\right)$

To find local maxima/minima, we use derivatives.

First derivative using Fundamental Theorem of Calculus

$f'(x)=\sqrt{x}\sin x$

Critical points occur where

$f'(x)=0$

Since $\sqrt{x}>0$ for $x>0$, this gives

$\sin x=0 \implies x=n\pi$

In the interval $\left(0,\frac{5\pi}{2}\right)$, the critical points are

$x=\pi,\;2\pi$

Determine nature of critical points

Since $\sqrt{x}>0$, the sign of $f'(x)$ is determined entirely by $\sin x$.

• For $x\in(0,\pi)$, $\sin x>0$, so $f'(x)>0$ Hence $f$ is increasing.

• For $x\in(\pi,2\pi)$, $\sin x<0$, so $f'(x)<0$ Hence $f$ is decreasing.

• For $x\in(2\pi,5\pi/2)$, $\sin x>0$, so $f'(x)>0$ Hence $f$ is increasing again.

Conclusion from sign change

• At $x=\pi$: $f'$ changes from positive to negative, so $f$ has a local maximum at $x=\pi$.
• At $x=2\pi$: $f'$ changes from negative to positive, so $f$ has a local minimum at $x=2\pi$.

Correct option

Therefore, $f$ has

$\text{local maximum at }\pi \text{ and local minimum at }2\pi$

So the correct choice is C.