If $\displaystyle I = \int_{0}^{\pi}\frac{\sin (884 x)\sin (1122x)}{2\sin x}dx$ and $\displaystyle J = \int_{0}^{1}\frac{x^{238}(x^{1768}-1)}{x^2-1}dx\;,$ Then $\displaystyle \frac{I}{J}$
$\bf{My\; Try::}$
$$\displaystyle \int_{0}^{\pi}\frac{\sin(884x)\times \sin(1122x)}{2\sin x}dx=\int_{0}^{\pi}\frac{\cos (238 x)- \cos(2006x)}{4\sin x}dx=\int_{0}^{\pi}\frac{\cos (238 x)}{4\sin x}dx-\int_{0}^{\pi}\frac{\cos(2006x)}{4\sin x}dx=I_{119} -I_{1003}$$
Where $$I_k=\int_{0}^{\pi}\frac{\cos (2k) x}{4\sin x}dx=\int_{0}^{\frac{\pi}{2}}\frac{\cos (2k) x}{4\sin x}+\frac{\cos (2k\pi-2k\pi x)}{4\sin (\pi-x)}dx=\int_{0}^{\frac{\pi}{2}}\frac{\cos (2k) x}{2\sin x}dx$$
$$\displaystyle \implies I_k-I_{k-1}=\int_{0}^{\frac{\pi}{2}}\frac{\cos (2k) x-\cos 2(k-1)x}{2\sin x}dx=-\int_{0}^{\frac{\pi}{2}}\frac{2\sin (2k-1) x\sin x}{2\sin x}dx$$
$$=-\int_{0}^{\frac{\pi}{2}}\sin (2k-1)dx=\biggl(\frac {\cos (2k-1)x}{2k-1}\biggr)_{0}^{\frac{\pi}{2}}=-\frac {1}{2k-1}$$
$$\implies I_{k-1}-I_{k}=\frac {1}{2k-1}$$
Now plug $k=120,121,122,\cdots,1003$ in the above expression and add all the equations we get
$$I_{119}-I_{1003}=\sum_{k=120}^{1003} \frac {1}{2k-1}$$
Hence $$\int_{0}^{\pi}\frac{\sin(884x)\times \sin(1122x)}{2\sin x}dx=\sum_{k=120}^{1003} \frac {1}{2k-1}$$
How can i solve Second Integral , Help required, Thanks
