I-n-0-pi-4-dx-cos-2n-1-x-to-show-that-n-N-2nI-n-2n-1-I-n-1-2-n-2-I-n-0-pi-4-1-cos-2n-1-x-1-cos-2-x-dx-

Question Number 149885 by puissant last updated on 08/Aug/21

$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}} \\ $$$${to}\:{show}\:{that}\:: \\ $$$$\forall\:{n}\in\mathbb{N}^{\ast} ,\:\mathrm{2}{nI}_{{n}} =\left(\mathrm{2}{n}−\mathrm{1}\right){I}_{{n}−\mathrm{1}} +\frac{\mathrm{2}^{{n}} }{\:\sqrt{\mathrm{2}}} \\ $$$$\left({I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}}×\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}\right){dx}\right)… \\ $$

Answered by Ar Brandon last updated on 08/Aug/21

$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sec}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}}{dx} \\ $$$$\:\:\:\:=\left[\frac{\mathrm{tan}{x}}{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\left(\mathrm{2}{n}−\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}{dx} \\ $$

Commented by Ar Brandon last updated on 08/Aug/21

$$\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x} \\ $$$$\mathrm{You}\:\mathrm{may}\:\mathrm{proceed}… \\ $$

Commented by puissant last updated on 27/Aug/21

$${oui}\:{en}\:{posant}\:{sin}^{\mathrm{2}} \left({x}\right)=\mathrm{1}−{cos}^{\mathrm{2}} \left({x}\right) \\ $$$${ca}\:{passe}\:{j}'{ai}\:{reussi}\:{a}\:{ecraser}\:{ca}\: \\ $$$${merci}\:{brooo}… \\ $$

Commented by Ar Brandon last updated on 08/Aug/21

$$\mathrm{D}'\mathrm{accord},\:\mathrm{ravi}\:!\:\mathrm{J}'\acute {\mathrm{e}tais}\:\mathrm{press}\acute {\mathrm{e}} \\ $$

Commented by puissant last updated on 08/Aug/21

$${confiance}.. \\ $$

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