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Question Number 101330 by 175 last updated on 01/Jul/20

$$Evaluate.$$$${\int }_{-\pi }^{\pi }{x}^9\cos xdx$$

Commented by mr W last updated on 01/Jul/20

$$thisisanoddfunction,since$$$$f(-x)=-f\left(x\right)$$$$$$$$foroddfunction:$$$${\int }_{-a}^{a}f\left(x\right)dx=0$$$$\Rightarrow {\int }_{-\pi }^{\pi }{x}^9\cos xdx=0$$

Commented by ajfour last updated on 02/Jul/20

$$wellwhatifI={\int }_0^{\pi }{x}^9\cos x?$$

Commented by mr W last updated on 02/Jul/20

$$I_{2n+1}=-(2n+1)[{\pi }^{2n}+\left(2n\right)I_{2n-1}]$$$$....$$

Commented by ajfour last updated on 02/Jul/20

$$thanksSir!$$

Commented by 1549442205 last updated on 02/Jul/20

$${\mathrm{F}}_{\mathrm{n}}=\int {\mathrm{x}}^{\mathrm{n}}\mathrm{dsinx}={\mathrm{x}}^{\mathrm{n}}\mathrm{sinx}-\int {\mathrm{nx}}^{\mathrm{n}-1}\mathrm{sinxdx}$$$$={\mathrm{x}}^{\mathrm{n}}\mathrm{sinx}+\mathrm{n}\int {\mathrm{x}}^{\mathrm{n}-1}\mathrm{dcosx}={\mathrm{x}}^{\mathrm{n}}\mathrm{sinx}+{\mathrm{nx}}^{\mathrm{n}-1}\mathrm{cosx}$$$$-\mathrm{n}(\mathrm{n}-1)\int {\mathrm{x}}^{\mathrm{n}-2}\mathrm{cosx}={\mathrm{x}}^{\mathrm{n}}\mathrm{sinx}+{\mathrm{nx}}^{\mathrm{n}-1}\mathrm{cosx}-\mathrm{n}(\mathrm{n}-1){\mathrm{F}}_{\mathrm{n}-2}$$$$\mathrm{Hence},{\mathrm{I}}_{\mathrm{n}}={\int }_0^{\pi }{\mathrm{x}}^{\mathrm{n}}\mathrm{cosxdx}={\mathrm{x}}^{\mathrm{n}}\mathrm{sinx}{\underset{0}{\mid }}^{\pi }+{\mathrm{nx}}^{\mathrm{n}-1}\mathrm{cosx}{\mid }_0^{\pi }-{\mathrm{nI}}_{\mathrm{n}-1}$$$${\mathbf{I}}_{\mathbf{n}}=-\mathbf{n}{\mathit{\pi }}^{\mathbf{n}-1}-\mathbf{n}(\mathbf{n}-1){\mathbf{I}}_{\mathbf{n}-2}$$

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