Find a real valued function $u(z)$ which is harmonic in an open disk and continuous up to boundary The radius of disk is said to be $R$, and on the boundary 
$$u(Re^{i\theta})=\dfrac{1}{2}(1+\cos^3\theta).$$ 
My thought was to integrate it using Poisson kernel, 
$$u(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{R^2-r^2}{R^2-2Rr\cos(\theta-t)+r^2}\frac{1}{2}(1+\cos^3\theta)dt.$$ 
However, I have trouble integrate this term. Can you suggest me another way of looking at this problem through complex analysis, partial differential equations or some skills in dealing with this integral? 

One trick is to recall the Poisson kernel on the unit disc can be written as
$$
P(z)=\mathrm{Re}\left(\frac{1+z}{1-z}\right)\text{ for }z\in\mathbb{D}
$$
and you can do the integral that way.  However, it is easier if you just "spot" the answer.
Rescale to $R=1$.  Then
$$
u(e^{i\theta})-\frac12=\frac{\cos^3\theta}{2}=\frac{2\cos(3\theta)+6\cos\theta}{16}
$$
so
$$
u(z)=\frac12+\frac{z^3+\bar{z}^3+3z+3\bar{z}}{16}=\mathrm{Re}\left(\frac12+\frac{z^3+3z}{8}\right).
$$
Rescale back to general $R$ gives
$$
u(z)=\mathrm{Re}\left(\frac12+\frac{z^3+3R^2z}{8R^3}\right)
$$
as the desired harmonic function
