definite integration

\hspace{-16}$For $\bf{(1)\;.......}$ Let $\bf{I = \int_{0}^{2008}x\left|\sin (\pi x)\right|dx}$\\\\\\ Let $\bf{\pi.x = t\;,}$ Then $\bf{dx = \frac{1}{\pi}dt}$ and Changing Limit, we Get\\\\\\ $\bf{I=\frac{1}{\pi^2}\int_{0}^{2008 \pi}t\left|\sin t\right|dt = \frac{2008}{\pi^2}\int_{0}^{\pi}t\cdot \left|\sin t\right|dt = \frac{2008}{\pi^2}\int_{0}^{\pi}t\cdot \sin tdt}$\\\\\\ (Bcz $\bf{\left|\sin t\right|}$ is a Periodic function with period $\bf{\pi}$ and in $\bf{0<t<\pi\;\;,\left|\sin t \right| = \sin t}$)\\\\\\ So $\bf{I = \frac{2008}{\pi^2}\int_{0}{\pi}t\cdot \sin tdt = \frac{2008}{\pi^2}\left(-t\cdot \cos t+\sin t\right)\Big|_0^1 = \frac{2008}{\pi^2}\cdot \pi =\frac{2008}{\pi}}$\\\\\\ So Expression $\bf{ \sqrt{\pi\cdot \left\{\int_{0}^{2008}x\left|\sin (\pi x)\right|dx\right\}}=\sqrt{\pi\cdot I} = \sqrt{\pi\cdot \frac{2008}{\pi}}=\sqrt{2008}}$

\hspace{-16}$For $\bf{(2)\;.......}$ Given $\bf{I_{n}=\int_{0}^{\frac{\pi}{2}}\frac{\sin (2n-1)x}{\sin x}dx}$ and $\bf{J_{n}=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2(nx)}{\sin^2 x}dx}$\\\\\\ Here $\bf{I_{n}=\int_{0}^{\frac{\pi}{2}}\frac{\sin (2n-1)x}{\sin x}dx.....................................(1)}$\\\\\\ Now Replace $\bf{n\rightarrow (n+1)\;,}$ We Get\\\\\\ So $\bf{I_{n+1}=\int_{0}^{\frac{\pi}{2}}\frac{\sin (2n+1)x}{\sin x}dx......................................(2)}$\\\\\\ Now from equation$\bf{(1)-(2)}$.......\\\\\\ So $\bf{I_{n+1}-I_{n}=\int_{0}^{\frac{\pi}{2}}\frac{\sin (2n+1)x-\sin (2n-1)x}{\sin x}dx}$\\\\\\ So $\bf{I_{n+1}-I_{n} = 2\int_{0}^{\frac{\pi}{2}}\frac{\cos\left\{(2n+1)x-(2n-1)x\right\}\cdot\sin\left\{(2n+1)x-(2n-1)x\right\}}{\sin x}dx}$\\\\\\ Now Using $\bf{\bullet\; \sin A-\sin B = 2\cos(A+B)\cdot \sin (A-B)}$\\\\\\So $\bf{I_{n+1}-I_{n} = 2\int_{0}^{\frac{\pi}{2}}=2\cos(2nx)=0\Rightarrow \boxed{\boxed{\bf{I_{n+1}=I_{n}}}}}$\\\\\\

\hspace{-16}$Similarly $\bf{J_{n}=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2(nx)}{\sin^2 x}dx.................................(3)}$\\\\\\ Now Replace $\bf{n\rightarrow (n+1)\;,}$ We Get\\\\\\ So $\bf{J_{n+1}=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2((n+1)x)}{\sin^2 x}dx...............................(4)}$\\\\\\ So from equations $\bf{(3)}$ and $\bf{(4).....}$\\\\\\ So $\bf{J_{n+1}-J_{n}=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2((n+1)x)-\sin^2(nx)}{\sin^2 x}dx}$\\\\\\ Now Using $\bf{\bullet\; \sin^2(A)-\sin^2(B)=\sin(A+B)\cdot \sin(A-B)}$\\\\\\ So $\bf{J_{n+1}-J_{n}=\int_{0}^{\frac{\pi}{2}}\frac{\sin (2n+1)x\cdot \sin x}{\sin^2 x}dx}$\\\\\\ So $\bf{J_{n+1}-J_{n}=\int_{0}^{\frac{\pi}{2}}\frac{\sin(2n+1)x}{\sin x}dx = I_{n+1}}}$\\\\\\ So $\bf{\bf{\boxed{\boxed{\bf{J_{n+1}-J_{n}=I_{n+1}}}}}$\\\\\\