For a symmetric thin airfoil, the camber line is the chord line:

$$z_c(x)=0,$$

so the thin-airfoil tangency equation is

$$U_\infty\alpha -rac1{2\pi}\operatorname{PV} \int_0^c\frac{\gamma(\xi)}{x-\xi},d\xi=0.$$

Map the chord to

$$\boxed{x=\frac c2(1-\cos\theta)},\qquad 0\le\theta\le\pi.$$

The leading edge is $\theta=0$ and the trailing edge is $\theta=\pi$.

Vortex-sheet solution

The Kutta-compatible solution is

$$\boxed{ \gamma(\theta) =2U_\infty\alpha \frac{1+\cos\theta}{\sin\theta} =2U_\infty\alpha\cot\frac\theta2 }.$$

At the trailing edge,

$$\gamma\to0,$$

so the Kutta condition is satisfied. At the ideal infinitely thin leading edge,

$$\gamma\to\infty.$$

That leading-edge singularity is a limitation of the zero-thickness model rather than a prediction of infinite velocity on a rounded real airfoil.

Verify the tangency equation

Let the integration coordinate be

$$\xi=\frac c2(1-\cos\varphi),$$

so

$$d\xi=\frac c2\sin\varphi,d\varphi.$$

For the proposed solution,

$$\gamma(\varphi)d\xi =U_\infty\alpha c(1+\cos\varphi),d\varphi.$$

Also,

$$x-\xi =\frac c2(\cos\varphi-\cos\theta).$$

Therefore

$$\operatorname{PV}\int_0^c \frac{\gamma(\xi)}{x-\xi},d\xi =2U_\infty\alpha\operatorname{PV} \int_0^\pi \frac{1+\cos\varphi} {\cos\varphi-\cos\theta},d\varphi.$$

Use

$$\frac{1+\cos\varphi}{\cos\varphi-\cos\theta} =1+ \frac{1+\cos\theta}{\cos\varphi-\cos\theta}.$$

For an interior point $0<\theta<\pi$, the standard principal-value identity

$$\operatorname{PV}\int_0^\pi \frac{d\varphi}{\cos\varphi-\cos\theta}=0$$

leaves

$$\operatorname{PV}\int_0^c \frac{\gamma(\xi)}{x-\xi},d\xi =2\pi U_\infty\alpha.$$

Hence

$$\frac1{2\pi}\operatorname{PV}\int_0^c \frac{\gamma(\xi)}{x-\xi},d\xi =U_\infty\alpha,$$

which satisfies the tangency equation exactly within thin-airfoil theory.

Distribution along the chord

Since

$$\cos\theta=1-2\frac xc,$$

and

$$\sin\theta =2\sqrt{\frac xc\left(1-\frac xc\right)},$$

we obtain

$$\boxed{ \gamma(x) =2U_\infty\alpha \sqrt{\frac{1-x/c}{x/c}} }.$$

The pressure-coefficient difference is

$$\boxed{ \Delta C_p(x) =C_{p,lower}-C_{p,upper} =4\alpha \sqrt{\frac{1-x/c}{x/c}} }.$$

It is strongest near the ideal leading edge and vanishes at the trailing edge.

Total circulation and lift

The positive sheet-strength integral is

$$G=\int_0^c\gamma(x),dx.$$

Using

$$dx=\frac c2\sin\theta,d\theta,$$

$$G =U_\infty c\alpha \int_0^\pi(1+\cos\theta),d\theta =\boxed{\pi U_\infty c\alpha}.$$

With counterclockwise circulation defined positive,

$$\Gamma=-G,$$

so

$$\boxed{\Gamma=-\pi U_\infty c\alpha}.$$

A positive angle of attack therefore selects clockwise bound circulation and upward lift under the established Kutta-Joukowski convention.

The lift per unit span is

$$L'=\rho U_\infty G =\pi\rho U_\infty^2c\alpha.$$

Define

$$C_L=\frac{L'}{\tfrac12\rho U_\infty^2c}.$$

Then

$$\boxed{C_L=2\pi\alpha},$$

with $\alpha$ in radians. Thus

$$\boxed{\frac{dC_L}{d\alpha}=2\pi\ \text{per radian}}.$$

Worked example

For

$$\alpha=5^\circ=5\frac\pi{180}\approx0.0873,$$

thin-airfoil theory predicts

$$C_L=2\pi(0.0873) \approx\boxed{0.548}.$$

At zero angle of attack, a symmetric thin airfoil has

$$C_L=0.$$

Scope

The $2\pi$ slope follows from a two-dimensional, incompressible, inviscid, attached-flow model at small angle and negligible thickness/camber to leading order. Finite thickness, viscosity, separation, compressibility, and finite span can all change the measured lift curve.

The result is therefore a foundational reference solution: it shows how a distributed vortex sheet satisfying tangency and the Kutta condition produces both the chordwise pressure loading and the integrated lift.