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Question Number 149739 by iloveisrael last updated on 07/Aug/21

Answered by Olaf_Thorendsen last updated on 07/Aug/21

$$\lambda ={\int }_0^{\pi }x{\sin }^{9}xdx\left(1\right)$$$$\lambda ={\int }_{\pi }^0(\pi -u){\sin }^9(\pi -u)(-du)$$$$\lambda ={\int }_0^{\pi }(\pi -u){\sin }^{9}udu\left(2\right)$$$$\left(1\right)+\left(2\right):2\lambda =\pi {\int }_0^{\pi }{\sin }^{9}xdx$$$$\lambda =\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{9}xdx+\frac{\pi }{2}{\int }_{\frac{\pi }{2}}^{\pi }{\sin }^{9}xdx$$$$\lambda =\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{9}xdx+\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\sin }^9(t+\frac{\pi }{2})dt$$$$\lambda =\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{9}xdx+\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\cos }^{9}tdt$$$$\lambda =\frac{\pi }{2}{\mathrm{W}}_9+\frac{\pi }{2}{\mathrm{W}}_9=\pi {\mathrm{W}}_9\left(\mathrm{Wallis}\right)$$$${\mathrm{W}}_{2p+1}=\frac{{(2^{p}p!)}^2}{(2p+1)!}$$$${\mathrm{W}}_9={\mathrm{W}}_{2\times 4+1}=\frac{{(2^{4}4!)}^2}{9!}=\frac{128}{315}$$$$\Rightarrow \lambda =\frac{128\pi }{315}$$

Answered by mathmax by abdo last updated on 07/Aug/21

$$\mathrm{\Psi }={\int }_0^{\pi }\mathrm{x}{\sin }^9\mathrm{x}\mathrm{dx}\Rightarrow \mathrm{\Psi }={\int }_0^{\pi }\mathrm{x}{\left(\frac{{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}{2\mathrm{i}}\right)}^9\mathrm{dx}$$$$=\frac{1}{{\left(2\mathrm{i}\right)}^9}{\int }_0^{\pi }\mathrm{x}\left(\sum _{\mathrm{k}=0}^9{\mathrm{C}}_9^{\mathrm{k}}{\left({\mathrm{e}}^{\mathrm{ix}}\right)}^{\mathrm{k}}{(-{\mathrm{e}}^{-\mathrm{ix}})}^{9-\mathrm{k}}\right)\mathrm{dx}$$$$=\frac{1}{{\left(2\mathrm{i}\right)}^9}\sum _{\mathrm{k}=0}^9{\mathrm{C}}_9^{\mathrm{k}}{(-1)}^{9-\mathrm{k}}{\int }_0^{\pi }\mathrm{x}{\mathrm{e}}^{\mathrm{ikx}}{\mathrm{e}}^{(\mathrm{k}-9)\mathrm{ix}}\mathrm{dx}$$$$=-\frac{1}{{\left(2\mathrm{i}\right)}^9}\sum _{\mathrm{k}=0}^9{(-1)}^{\mathrm{k}}{\mathrm{C}}_9^{\mathrm{k}}{\mathrm{u}}_{\mathrm{k}}$$$${\mathrm{u}}_{\mathrm{k}}={\int }_0^{\pi }\mathrm{x}{\mathrm{e}}^{(2\mathrm{k}-9)\mathrm{ix}}\mathrm{dx}={\left[\frac{\mathrm{x}}{(2\mathrm{k}-9)\mathrm{i}}{\mathrm{e}}^{(2\mathrm{k}-9)\mathrm{ix}}\right]}_0^{\pi }-{\int }_0^{\pi }\frac{1}{(2\mathrm{k}-9)\mathrm{i}}{\mathrm{e}}^{(2\mathrm{k}-9)\mathrm{ix}}\mathrm{dx}$$$$=\frac{\pi }{(2\mathrm{k}-9)\mathrm{i}}{\mathrm{e}}^{(2\mathrm{k}-9)\mathrm{i}\pi }-\frac{1}{{\left((2\mathrm{k}-9)\mathrm{i}\right)}^2}{\left[{\mathrm{e}}^{(2\mathrm{k}-9)\mathrm{ix}}\right]}_0^{\pi }$$$$=-\frac{\pi }{(2\mathrm{k}-9)\mathrm{i}}+\frac{1}{{(2\mathrm{k}-9)}^2}(-2)\Rightarrow$$$$\mathrm{\Psi }=-\frac{1}{{\left(2\mathrm{i}\right)}^9}\sum _{\mathrm{k}=0}^9{(-1)}^{\mathrm{k}}{\mathrm{C}}_9^{\mathrm{k}}(-\frac{\pi }{(2\mathrm{k}-9)\mathrm{i}}-\frac{2}{{(2\mathrm{k}-9)}^2})....$$

Answered by gsk2684 last updated on 07/Aug/21

$$\lambda +\lambda =\underset{0}{\stackrel{\pi }{\int }}x\sin {}^{9}xdx+\underset{0}{\stackrel{\pi }{\int }}(\pi -x)\sin {}^9(\pi -x)dx$$$$2\lambda =\underset{0}{\stackrel{\pi }{\int }}x\sin {}^{9}xdx+\underset{0}{\stackrel{\pi }{\int }}(\pi -x)\sin {}^{9}xdx$$$$2\lambda =\underset{0}{\stackrel{\pi }{\int }}x\sin {}^{9}xdx+\pi \underset{0}{\stackrel{\pi }{\int }}\sin {}^{9}xdx-\underset{0}{\stackrel{\pi }{\int }}x\sin {}^{9}xdx$$$$2\lambda =\pi \underset{0}{\stackrel{\pi }{\int }}\sin {}^{9}xdx$$$$2\lambda =2\pi \underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\sin {}^{9}xdx$$$$\lambda =\pi \frac{8}{9}\frac{6}{7}\frac{4}{5}\frac{2}{3}1$$$$$$

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