The first quadrant restricts us to the domain [tex]0\le\theta\le\dfrac\pi2[/tex]. So the area of the region, call it [tex]R[/tex], is
[tex]\displaystyle\iint_R\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=2\sin\theta+4\cos\theta}r\,\mathrm dr\,\mathrm d\theta=\frac12\int_{\theta=0}^{\theta=\pi/2}(2\sin\theta+4\cos\theta)^2\,\mathrm d\theta[/tex]
Expanding the integrand yields
[tex](2\sin\theta+4\cos\theta)^2=4\sin^2\theta+16\sin\theta\cos\theta+16\cos^2\theta=4+8\sin\theta\cos\theta+12\cos^2\theta[/tex]
Apply the double angle identities:
[tex]\sin2\theta=2\sin\theta\cos\theta\implies16\sin\theta\cos\theta=8\sin2\theta[/tex]
[tex]\cos^2\theta=\dfrac{1+\cos2\theta}2[/tex]
So the integral is
[tex]\displaystyle\int_0^{\pi/2}(5+4\sin2\theta+3\cos2\theta)\,\mathrm d\theta[/tex]
[tex]=5\theta-2\cos2\theta+\dfrac32\sin2\theta\bigg|_0^{\pi/2}[/tex]
[tex]=\left(\dfrac{5\pi}2-2\cos\pi+\dfrac32\sin\pi\right)-\left(0-2\cos0+0\right)=\dfrac{5\pi}2+4[/tex]
so the answer is C.
