\(\sin x\ge0\) for all \(x\in[0,\pi]\), while \(\cos x>0\) for \(0 and \(\cos x<0\) for \(\dfrac\pi2. So you already know that \(4\sin x>10\cos x\) over the second half of the interval.
In the first half, \(4\sin x\) is an increasing function, from \(4\sin0=0\) to \(4\sin\dfrac\pi2=4\). Meanwhile \(10\cos x\) is a decreasing function, from \(10\cos0=10\) to \(10\cos\dfrac\pi2=0\). Therefore there must be a point between 0 and \(\dfrac\pi2\) where the two curves intersect. So we find it:
\(4\sin x=10\cos x\implies \tan x=\dfrac{10}4=\dfrac52\implies x=\arctan\dfrac52\approx1.1903\)
So we know that \(10\cos x>4\sin x\) in \(\left[0,\arctan\dfrac52\right)\), and \(4\sin x>10\cos x\) in \(\left(\arctan\dfrac52,\pi\right]\).
The area between the curves is then
\(\displaystyle\int_0^{\arctan(5/2)}(10\cos x-4\sin x)\,\mathrm dx+\int_{\arctan(5/2)}^\pi(4\sin x-10\cos x)\,\mathrm dx=4\sqrt{29}\)
