Definite integral using properties

$$I=4pi ln 14+int_{0}^{4pi} ln sin (x+alpha)~ dx$$ Let $x+alpha=t$, then $$I=4piln 14+ int_{alpha}^{4pi+alpha} log sin t dt$$ Since integrand is periodic funxtion eyj period $2pi$ then $$I=4pi ln 14+2int_{0}^{2pi} ln sin t ~ dt~~~~(1)$$ Use $$int_{0}^{2a} f(x) dx= int_{0}^{a} [f(x)+f(2a-x)] dx ~~~~~~~(2)$$ Using (1), we get $$int_{0}^{2pi} ln sin t dt=int_{0}^{pi} [ln sin t+ ln (-sin t)]_~dt= ipi^2+2int_{0}^{pi} ln sin t~ dt$$ Using (1) again, we get $$K=int_{0}^{pi} ln sin t~ dt= 2int_{0}^{pi/2} sin t ~dt= 2 J$$ $$int_{0}^{a} f(x) dx=int_{0}^{a} f(a-x) dx~~~~~(3)$$

$$J=int_{0}^{pi/2} ln sin t dt implies int_{0}^{pi/2} ln cos t ~dt implies 2J=int_{0}^{pi/2} ln (frac{sin 2t}{2}) dt=frac{1}{2}int_{0}^{pi} ln sin u~du-frac{pi}{2}ln 2$$ $$2J=K/2-frac{pi}{2}ln 2 implies J=-frac{pi}{2}ln 2$$

Finally, using there results in (1), we get $$I=4piln 14+2ipi^2-4pi ln 2=4pi ln 7+2ipi^2$$