Evaluate-pi-pi-x-9-cos-x-dx-

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

$${Evaluate}. \\ $$$$\int_{−\pi} ^{\pi} {x}^{\mathrm{9}} \mathrm{cos}\:{x}\:{dx} \\ $$

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

this is an odd function, since f(−x)=−f(x) for odd function: ∫_(−a) ^a f(x)dx=0 ⇒∫_(−π) ^π x^9 cos x dx=0

$${this}\:{is}\:{an}\:{odd}\:{function},\:{since} \\ $$$${f}\left(−{x}\right)=−{f}\left({x}\right) \\ $$$$ \\ $$$${for}\:{odd}\:{function}: \\ $$$$\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}=\mathrm{0} \\ $$$$\Rightarrow\int_{−\pi} ^{\pi} {x}^{\mathrm{9}} \mathrm{cos}\:{x}\:{dx}=\mathrm{0} \\ $$

Commented by ajfour last updated on 02/Jul/20

$${well}\:{what}\:{if}\:\:{I}=\:\int_{\mathrm{0}} ^{\:\:\pi} {x}^{\mathrm{9}} \mathrm{cos}\:{x}\:? \\ $$

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

I_(2n+1) =−(2n+1)[π^(2n) +(2n)I_(2n−1) ] ....

$${I}_{\mathrm{2}{n}+\mathrm{1}} =−\left(\mathrm{2}{n}+\mathrm{1}\right)\left[\pi^{\mathrm{2}{n}} +\left(\mathrm{2}{n}\right){I}_{\mathrm{2}{n}−\mathrm{1}} \right] \\ $$$$…. \\ $$

Commented by ajfour last updated on 02/Jul/20

$${thanks}\:{Sir}! \\ $$

Commented by 1549442205 last updated on 02/Jul/20

F_n =∫x^n dsinx=x^n sinx−∫nx^(n−1) sinxdx =x^n sinx+n∫x^(n−1) dcosx=x^n sinx+nx^(n−1) cosx −n(n−1)∫x^(n−2) cosx=x^n sinx+nx^(n−1) cosx−n(n−1)F_(n−2) Hence,I_n =∫_0 ^π x^n cosxdx=x^n sinx∣^π _0 +nx^(n−1) cosx∣_0 ^π −nI_(n−1) I_n =−n𝛑^(n−1) −n(n−1)I_(n−2)

$$\mathrm{F}_{\mathrm{n}} =\int\mathrm{x}^{\mathrm{n}} \mathrm{dsinx}=\mathrm{x}^{\mathrm{n}} \mathrm{sinx}−\int\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \mathrm{sinxdx} \\ $$$$=\mathrm{x}^{\mathrm{n}} \mathrm{sinx}+\mathrm{n}\int\mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{dcosx}=\mathrm{x}^{\mathrm{n}} \mathrm{sinx}+\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \mathrm{cosx} \\ $$$$−\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\int\mathrm{x}^{\mathrm{n}−\mathrm{2}} \mathrm{cosx}=\mathrm{x}^{\mathrm{n}} \mathrm{sinx}+\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \mathrm{cosx}−\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\mathrm{F}_{\mathrm{n}−\mathrm{2}} \\ $$$$\mathrm{Hence},\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\pi} \mathrm{x}^{\mathrm{n}} \mathrm{cosxdx}=\mathrm{x}^{\mathrm{n}} \mathrm{sinx}\underset{\mathrm{0}} {\mid}^{\pi} +\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \mathrm{cosx}\mid_{\mathrm{0}} ^{\pi} −\mathrm{nI}_{\mathrm{n}−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{I}}_{\boldsymbol{\mathrm{n}}} =−\boldsymbol{\mathrm{n}\pi}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} −\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\boldsymbol{\mathrm{I}}_{\boldsymbol{\mathrm{n}}−\mathrm{2}} \\ $$

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