Calculate the line integral $I(a,b)=\int_{x^2+y^2=R^2}\limits\ln\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} ds\quad(a^2+b^2\ne R^2).$ Calculate the line integral $$I(a,b)=\int_{x^2+y^2=R^2}\limits\ln\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} ds\quad(a^2+b^2\ne R^2).$$
The parametrized integral path can be given as
$$x=R\cos t,y=R\sin t,t\in[0,2\pi].$$
Then I get into trouble when computing the integral
$$-\frac{1}{2}\int_{0}^{2\pi}R\ln(a^2+b^2+R^2-2aR\cos t-2bR\sin t) dt.$$
Should I take the derivative with respect to $R$ first? Or apply integration by parts?
Update: I find the offered answer is
$$I(a,b)=-2\pi R\ln\max\{R,\sqrt{a^2+b^2}\}$$
and I have examined its correctness.

Hint : $$W=C\sin(t)+D\cos(t)=(\sqrt{C^2+D^2})(\frac{C}{\sqrt{C^2+D^2}}\sin(t)+\frac{D}{\sqrt{C^2+D^2}}\cos(t))$$
$$W=\sqrt{C^2+D^2}\sin(t+\phi)=E\sin(t+\phi)$$
with $\cos(\phi)=\frac{C}{\sqrt{C^2+D^2}}$ and $ \sin(\phi)=\frac{D}{\sqrt{C^2+D^2}}$
so $$I=\int_0^{2\pi} \ln(A+C\sin(t)+B\cos(t))\text{d}t=\int_0^{2\pi} \ln(A+E\sin(t+\phi))\text{d}t$$
here a formal calculation software can complete the calculation
