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

I_n =∫_0 ^(π/4) (dx/(cos^(2n+1) x))  to show that :  ∀ n∈N^∗ , 2nI_n =(2n−1)I_(n−1) +(2^n /( (√2)))  (I_n =∫_0 ^(π/4) ((1/(cos^(2n−1) x))×(1/(cos^2 x)))dx)...

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

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

I_n =∫_0 ^(π/4) (dx/(cos^(2n+1) x))=∫_0 ^(π/4) ((sec^2 x)/(cos^(2n−1) x))dx      =[((tanx)/(cos^(2n−1) x))]_0 ^(π/4) −(2n−1)∫_0 ^(π/4) ((sin^2 x)/(cos^(2n+1) x))dx

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

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

sin^2 x=1−cos^2 x  You may proceed...

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

Commented by puissant last updated on 27/Aug/21

oui en posant sin^2 (x)=1−cos^2 (x)  ca passe j′ai reussi a ecraser ca   merci brooo...

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

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

D′accord, ravi ! J′e^� tais presse^�

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

Commented by puissant last updated on 08/Aug/21

confiance..

$${confiance}.. \\ $$

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