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Question Number 115014 by mathdave last updated on 23/Sep/20

$$if{I}_n={\int }_x^{\frac{\pi }{2}}x{\cos }^{n}xdx,wheren\succ 1show$$$$that{I}_n=\frac{n(n-1)I_{n-2}-1}{n^2}andthen$$$$evaluate{\int }_x^{\frac{\pi }{2}}x{\cos }^{8}xdx$$

Answered by Olaf last updated on 23/Sep/20

$$$$$${\mathrm{I}}_n={\int }_0^{\frac{\pi }{2}}x{\cos }^{n}xdx$$$${\mathrm{I}}_n-{\mathrm{I}}_{n+2}={\int }_0^{\frac{\pi }{2}}x{\cos }^{n}x(1-{\cos }^{2}x)dx$$$${\mathrm{I}}_n-{\mathrm{I}}_{n+2}=-{\int }_0^{\frac{\pi }{2}}\left(x\sin x\right)(-\sin x{\cos }^{n}x)dx$$$${\mathrm{I}}_n-{\mathrm{I}}_{n+2}=-{\left[x\sin x\frac{{\cos }^{n+1}x}{n+1}\right]}_0^{\frac{\pi }{2}}$$$$+{\int }_0^{\frac{\pi }{2}}(\sin x+x\cos x)\frac{{\cos }^{n+1}x}{n+1}dx$$$${\mathrm{I}}_n-{\mathrm{I}}_{n+2}=-{\int }_0^{\frac{\pi }{2}}(-\sin x\frac{{\cos }^{n+1}x}{n+1})dx$$$$+\frac{1}{n+1}{\mathrm{I}}_{n+2}$$$${\mathrm{I}}_n-{\mathrm{I}}_{n+2}=-{\left[\frac{{\cos }^{n+2}x}{(n+1)(n+2)}\right]}_0^{\frac{\pi }{2}}+\frac{1}{n+1}{\mathrm{I}}_{n+2}$$$${\mathrm{I}}_n=\frac{1}{(n+1)(n+2)}+{\mathrm{I}}_{n+2}+\frac{1}{n+1}{\mathrm{I}}_{n+2}$$$${\mathrm{I}}_n=\frac{n+2}{n+1}{\mathrm{I}}_{n+2}+\frac{1}{(n+1)(n+2)}$$$${\mathrm{I}}_{n+2}=\frac{n+1}{n+2}{\mathrm{I}}_n-\frac{1}{{(n+2)}^2}\left(1\right)$$$${\mathrm{I}}_n=\frac{n(n-1){\mathrm{I}}_{n-2}-1}{n^2}$$$${\mathrm{I}}_0={\int }_0^{\frac{\pi }{2}}xdx={\left[\frac{x^2}{2}\right]}_0^{\frac{\pi }{2}}=\frac{{\pi }^2}{8}$$$$\mathrm{With}\left(1\right):{\mathrm{I}}_2=\frac{1}{2}\left(\frac{{\pi }^2}{8}\right)-\frac{1}{4}=\frac{{\pi }^2}{16}-\frac{1}{4}$$$${\mathrm{I}}_4=\frac{3}{4}(\frac{{\pi }^2}{16}-\frac{1}{4})-\frac{1}{16}=\frac{3{\pi }^2}{64}-\frac{1}{4}$$$${\mathrm{I}}_6=\frac{5}{6}(\frac{3{\pi }^2}{64}-\frac{1}{4})-\frac{1}{36}=\frac{5{\pi }^2}{128}-\frac{17}{72}$$$${\mathrm{I}}_8=\frac{7}{8}(\frac{5{\pi }^2}{128}-\frac{17}{72})-\frac{1}{64}=\frac{35{\pi }^2}{1024}-\frac{2}{9}$$$${\mathrm{I}}_8={\int }_0^{\frac{\pi }{2}}x{\cos }^{8}xdx=\frac{35{\pi }^2}{1024}-\frac{2}{9}$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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