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Question Number 34901 by rishabh last updated on 12/May/18

∫_(−π/2) ^(+π/2) (√(cos^(2n−1) x−cos^(2n+1) x))dx  =[−((2cos^((2n+1)/2) x)/(2n+1))]_(−π/2) ^(+π/2) =0?  What is the mistake in above?  ∫_(−π/2) ^(+π/2) (√(cos^(2n−1) x−cos^(2n+1) x))dx  =2∫_0 ^(π/2) (√(cos^(2n−1) x−cos^(2n+1) x))dx  =(4/(2n+1)) (this is correct answer)

π/2+π/2cos2n1xcos2n+1xdx=[2cos2n+12x2n+1]π/2+π/2=0?Whatisthemistakeinabove?π/2+π/2cos2n1xcos2n+1xdx=20π/2cos2n1xcos2n+1xdx=42n+1(thisiscorrectanswer)

Commented by rishabh last updated on 13/May/18

What is the mistake in first method  which gives answer 0?

Whatisthemistakeinfirstmethodwhichgivesanswer0?

Commented by math khazana by abdo last updated on 13/May/18

 I = 2 ∫_0 ^(π/2)  (√(cos^(2n−1) x −cos^(2n+1) x)) dx    chang. cos^(2n+1) x =t  ⇒cosx =t^(1/(2n+1))  ⇒  x =arcos(t^(1/(2n+1)) )   cos^(2n−1) x= cos^(2n+1 −2) =(t/(cos^2 x))  =  (t/t^(2/(2n+1)) ) ⇒  I = 2 ∫_1 ^0  (√(  (t/t^(2/(2n+1)) )−t))     (1/(2n+1)) t^((1/(2n+1))−1)  ((−dt)/(√(1−t^(2/(2n+1)) )))  = (2/(2n+1)) ∫_0 ^1     (√t) (√((1−t^(2/(2n+1)) )/t^(2/(2n+1)) ))  .t^((1/(2n+1))−1)    (dt/(√(1−t^(2/(2n+1)) )))  = (2/(2n+1)) ∫_0 ^1     (√t) t^(1/(2n+1))   t^( (1/(2n+1)) −1) dt  = (2/(2n+1)) ∫_0 ^1   (√t)  t^((2/(2n+1))−1) dt   ((√t)=x)  = (2/(2n+1)) ∫_0 ^1  x . (x^2 )^((2/(2n+1))−1 ) dx  = (2/(2n+1)) ∫_0 ^1    x^(4/(2n+1)) dx  = (2/(2n+1))[  (1/((4/(2n+1))+1)) x^((4/(2n+1))+1) ]_0 ^1   = (2/(2n+1))  (1/((2n+5)/(2n+1))) =(2/(2n+5))

I=20π2cos2n1xcos2n+1xdxchang.cos2n+1x=tcosx=t12n+1x=arcos(t12n+1)cos2n1x=cos2n+12=tcos2x=tt22n+1I=210tt22n+1t12n+1t12n+11dt1t22n+1=22n+101t1t22n+1t22n+1.t12n+11dt1t22n+1=22n+101tt12n+1t12n+11dt=22n+101tt22n+11dt(t=x)=22n+101x.(x2)22n+11dx=22n+101x42n+1dx=22n+1[142n+1+1x42n+1+1]01=22n+112n+52n+1=22n+5

Commented by math khazana by abdo last updated on 13/May/18

error in the final lines  I = (2/(2n+1)) ∫_0 ^1   x (x^2 )^((2/(2n+1))−1)  (2xdx)  I = (4/(2n+1)) ∫_0 ^1    x^2  x^((4/(2n+1)) −2)  dx  = (4/(2n+1)) ∫_0 ^1    x^(4/(2n+1)) dx  = (4/(2n+1))[ (1/((4/(2n+1)) +1)) x^((4/(2n+1)) +1) ]_0 ^1   = (4/(2n+1)) .(1/((2n+5)/(2n+1))) = (4/(2n+5))  I = (4/(2n+5)) .

errorinthefinallinesI=22n+101x(x2)22n+11(2xdx)I=42n+101x2x42n+12dx=42n+101x42n+1dx=42n+1[142n+1+1x42n+1+1]01=42n+1.12n+52n+1=42n+5I=42n+5.

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