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Evaluate-lim-n-0-1-x-n-cos-x-dx-




Question Number 382 by novrya last updated on 25/Jan/15
Evaluate   lim_(n⇒∞)  ∫_0 ^1  (x^n /(cos x)) dx
Evaluatelimn10xncosxdx
Commented by 123456 last updated on 26/Dec/14
0≤x≤1⇒0≤x^n ≤x≤1  0≤x≤1⇒cos 1≤cos x≤1
0x10xnx10x1cos1cosx1
Commented by 123456 last updated on 26/Dec/14
f=(x^n /(cos x))  (∂f/∂x)=((nx^(n−1) cos x−x^n sin x)/(cos^2 x))  (∂f/∂x)=0(x=0)  f=0(x=0)  (∂f/∂x)=((ncos 1−sin 1)/(cos^2 1))≥^? 0,n∈N^∗ (x=1)  f=(1/(cos 1))(x=1)
f=xncosxfx=nxn1cosxxnsinxcos2xfx=0(x=0)f=0(x=0)fx=ncos1sin1cos21?0,nN(x=1)f=1cos1(x=1)
Commented by 123456 last updated on 26/Dec/14
0≤^? ∫_0 ^1 fdx≤^? (1/(cos 1))
0?10fdx?1cos1
Answered by prakash jain last updated on 27/Dec/14
Trying with series expansion of sec x for ∣x∣<(π/2)  ∫ (x^n /(cos x))dx=∫x^n sec xdx  =∫x^n [1+(1/2)x^2 +(5/(24))x^4 +((61)/(720))x^6 +...]dx  =(x^(n+1) /(n+1))+(1/2)∙(x^(n+3) /(n+3))+(5/(24))∙(x^(n+5) /(n+5))+..  With x=1 and n→∞ the given limit  is 0.
Tryingwithseriesexpansionofsecxforx∣<π2xncosxdx=xnsecxdx=xn[1+12x2+524x4+61720x6+]dx=xn+1n+1+12xn+3n+3+524xn+5n+5+..Withx=1andnthegivenlimitis0.