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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/2cos2n−1x−cos2n+1xdx=[−2cos2n+12x2n+1]−π/2+π/2=0?Whatisthemistakeinabove?∫−π/2+π/2cos2n−1x−cos2n+1xdx=2∫0π/2cos2n−1x−cos2n+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=2∫0π2cos2n−1x−cos2n+1xdxchang.cos2n+1x=t⇒cosx=t12n+1⇒x=arcos(t12n+1)cos2n−1x=cos2n+1−2=tcos2x=tt22n+1⇒I=2∫10tt22n+1−t12n+1t12n+1−1−dt1−t22n+1=22n+1∫01t1−t22n+1t22n+1.t12n+1−1dt1−t22n+1=22n+1∫01tt12n+1t12n+1−1dt=22n+1∫01tt22n+1−1dt(t=x)=22n+1∫01x.(x2)22n+1−1dx=22n+1∫01x42n+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+1∫01x(x2)22n+1−1(2xdx)I=42n+1∫01x2x42n+1−2dx=42n+1∫01x42n+1dx=42n+1[142n+1+1x42n+1+1]01=42n+1.12n+52n+1=42n+5I=42n+5.

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