Why is the coefficient of drag for straight wings at supersonic speeds lower than for swept wings?
Mark is right when he says that there is no induced drag at supersonic speeds, but it is an invitation for misunderstandings. Induced drag is replaced by lift wave drag, and all what happens is that aerodynamicists choose to use two different names for basically the same effect: Air gets pushed down.

As usual when I post a long answer, I was not quite happy with the existing answer(s). Now I have some time and try to give a better answer.

First, why does the straight wing work better only at very high Mach numbers (> 2.0)? Because at lower supersonic speed a swept wing gives overall better performance. The sweep angle $\varphi_0$ must be high enough to allow for a subsonic leading edge (Mach &lt; $\frac{1}{cos\varphi_0}$). Then the flow around the leading edge is subsonic and creates a suction area when accelerating around the nose contour. This suction helps to reduce drag - after all, this same suction is why a subsonic airfoil in inviscid flow has no drag. Edward C. Polhamus did a lot of research on this at NACA Langley and published several papers with equations for calculating the suction force.

Once you fly faster than Mach 2, the sweep angle for a subsonic leading edge rapidly gets too high for acceptable subsonic flight, and an unswept wing becomes the better alternative since you need to accept a supersonic leading edge. Examples are the wing of the F-104 or the canard of the XB-70.

Now for the airfoil drag at supersonic speed. Because it is easiest to explain, I select a rhombic cross section:


Rhombic airfoil in supersonic flow at zero angle of attack (own work). The plus sign denotes higher pressure, the minus sign lower pressure than ambient. By selecting a rhombic airfoil, the flow is very easy to determine because the pressure only changes when the local contour gradient changes. The two compression shocks create the typical sonic boom when arriving at the ground. Note that this airfoil already creates pressure drag even at zero lift. Any airfoil thickness greater than zero and any airfoil camber will cause this type of drag where the forward-facing areas see higher pressure and the rear-facing areas experience suction. This type of drag is called wave drag. It can only be minimized by minimizing the relative thickness of whatever is supposed fly at supersonic speed.

When the angle of attack is increased, this airfoil begins to create lift. Now the compression by the lower forward shock becomes stronger and that by the upper forward shock becomes weaker. The expansion fan is the same again on both sides, so the upper rear half experiences less pressure than the lower rear half. I tried to symbolize this by the amount of plus and minus signs:


Note that the pressure difference is constant over chord, so the center of pressure is at 50% of chord length. Note also that the lift vector is perpendicular to the chord line. Since lift is defined as the force perpendicular to the direction of the undisturbed air, supersonic lift always carries a drag component which is proportional to the angle of attack - there is no suction at the nose to alleviate this! The wave drag of the airfoil at zero angle of attack still comes on top, so we have a shape-dependent wave drag and a lift-dependent wave drag component. This lift-dependent wave drag replaces the induced drag of subsonic speeds. If we compare the magnitude of both, we find:

Subsonic: $c_{Di} = \frac{c^2_L}{\pi\cdot AR\cdot\epsilon}$

Supersonic formula for 2D flow: $c_{{DW}_L} = c_L\cdot\alpha$

This doesn't look so similar, so let's now express the angle of attack $\alpha$ by the lift coefficient divided by the lift curve slope:

$$\alpha = \frac{c_L}{c_{L\alpha}} = \frac{c_L}{\frac{4}{\sqrt{Ma^2-1}}\cdot\left(1 - \frac{\lambda}{2\cdot AR\cdot\sqrt{Ma^2-1}}\right)}$$

and the lift wave drag component becomes $c_{{DW}_L} = \frac{c^2_L}{\frac{4}{\sqrt{Ma^2-1}}\cdot\left(1 - \frac{\lambda}{2\cdot AR\cdot\sqrt{Ma^2-1}}\right)}$

Now let's compare the F-104 wing, which has an aspect ratio $AR$ of 2.45 and a taper ratio $\lambda$ of 0.385: If we plug in the parameters and adjust $\epsilon$ such that both sub- and supersonic lift-dependent drag coefficients agree, $\epsilon$ would need to be 0.89 at Mach 1.2, 0.58 at Mach 1.4 and 0.31 at Mach 2.0. The dramatic rise of lift wave drag over Mach is caused by the reduction in the lift curve slope over Mach.

For slender bodies the lift curve slope is $c_{L\alpha} = \frac{\pi\cdot AR}{2}$ and the lift wave drag component becomes $c_{{DW}_L} = 2\cdot\frac{c^2_L}{\pi\cdot AR}$. For slender bodies the supersonic $\epsilon$ is 0.5, regardless of Mach.

The important conclusions from this for wing selection are:


Sweep does no longer help once the leading edge is supersonic.
Lift wave drag continues where induced drag drops off. Lift always causes drag.
For a supersonic wing the aspect ratio is of minor importance.


Now back to the original question: Once the leading edge is supersonic, sweep is no longer helpful. Now the best wing is straight, because it will need the lowest wing area to create the required lift at subsonic speed. At supersonic speed its lower area will translate into lower friction drag, making it better than comparable delta or swept wings.