Find the value of the following integral.$$\int_{0}^{4\pi} \ln|13\sin x+3\sqrt3 \cos x|\;\rm dx$$
My attempt: Using the properties of definite integral, I converted it to this integral. $$4\int_{0}^{\pi} \ln|14\sin(x+\arctan(\frac{3\sqrt3}{13}))|\;\rm dx$$Please tell me how to proceed further.
$$I=4\pi \ln 14+\int_{0}^{4\pi} \ln \sin (x+\alpha)~ dx$$ Let $x+\alpha=t$, then $$I=4\pi\ln 14+ \int_{\alpha}^{4\pi+\alpha} \log \sin t dt$$ Since integrand is periodic funxtion eyj period $2\pi$ then $$I=4\pi \ln 14+2\int_{0}^{2\pi} \ln \sin t ~ dt~~~~(1)$$ Use $$\int_{0}^{2a} f(x) dx= \int_{0}^{a} [f(x)+f(2a-x)] dx ~~~~~~~(2)$$ Using (1), we get $$\int_{0}^{2\pi} \ln \sin t dt=\int_{0}^{\pi} [\ln \sin t+ \ln (-\sin t)]_~dt= i\pi^2+2\int_{0}^{\pi} \ln \sin t~ dt$$ Using (1) again, we get $$K=\int_{0}^{\pi} \ln \sin t~ dt= 2\int_{0}^{\pi/2} \sin t ~dt= 2 J$$ $$\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx~~~~~(3)$$
$$J=\int_{0}^{\pi/2} \ln \sin t dt \implies \int_{0}^{\pi/2} \ln \cos t ~dt \implies 2J=\int_{0}^{\pi/2} \ln (\frac{\sin 2t}{2}) dt=\frac{1}{2}\int_{0}^{\pi} \ln \sin u~du-\frac{\pi}{2}\ln 2$$ $$2J=K/2-\frac{\pi}{2}\ln 2 \implies J=-\frac{\pi}{2}\ln 2$$
Finally, using there results in (1), we get $$I=4\pi\ln 14+2i\pi^2-4\pi \ln 2=4\pi \ln 7+2i\pi^2$$