Answer
$f(4)=\frac{\pi}{2}$
Work Step by Step
$x\sin \pi x=\int_0^{x^2}f(t)dt$ (Differentiate with respect to $x$)
$\frac{d}{dx}(x\sin \pi x)=\frac{d}{dx}\int_0^{x^2}f(t)dt$ (Apply the Product Rule)
$\frac{d}{dx}(x)\cdot \sin (\pi x)+x\cdot \frac{d}{dx}(\sin \pi x)=\frac{d}{dx}\int_0^{x^2}f(t)dt$ (Apply the Chain Rule)
$1\cdot \sin (\pi x)+x\cdot \pi (\cos \pi x)=\frac{d}{dx}(x^2)\cdot \frac{d}{d(x^2)}\int_0^{x^2}f(t)dt$ (Use the Fundamental Theorem of Calculus in Part 1)
$\sin (\pi x)+\pi x\cos \pi x=2x\cdot f(x^2)$ (Replace $x$ by $2$)
$\sin (2\pi)+2\pi \cos (2\pi)=2\cdot 2\cdot f(2^2)$ (Simplify)
$0+2\pi \cdot 1=4f(4)$
$2\pi =4f(4)$
$f(4)=\frac{\pi}{2}$
