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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)

$$\int_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} \sqrt{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}−\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}{dx} \\ $$$$=\left[−\frac{\mathrm{2cos}^{\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}} {x}}{\mathrm{2}{n}+\mathrm{1}}\right]_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} =\mathrm{0}? \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{above}? \\ $$$$\int_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} \sqrt{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}−\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \sqrt{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}−\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}{dx} \\ $$$$=\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:\left(\mathrm{this}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{answer}\right) \\ $$

Commented by rishabh last updated on 13/May/18

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

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{first}\:\mathrm{method} \\ $$$$\mathrm{which}\:\mathrm{gives}\:\mathrm{answer}\:\mathrm{0}? \\ $$

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))

$$\:{I}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}\:−{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}\:{dx}\:\: \\ $$$${chang}.\:{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}\:={t}\:\:\Rightarrow{cosx}\:={t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} \:\Rightarrow \\ $$$${x}\:={arcos}\left({t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} \right)\:\:\:{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}=\:{cos}^{\mathrm{2}{n}+\mathrm{1}\:−\mathrm{2}} =\frac{{t}}{{cos}^{\mathrm{2}} {x}} \\ $$$$=\:\:\frac{{t}}{{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }\:\Rightarrow \\ $$$${I}\:=\:\mathrm{2}\:\int_{\mathrm{1}} ^{\mathrm{0}} \:\sqrt{\:\:\frac{{t}}{{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }−{t}}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}} \:\frac{−{dt}}{\sqrt{\mathrm{1}−{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\sqrt{{t}}\:\sqrt{\frac{\mathrm{1}−{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }{{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }}\:\:.{t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}} \:\:\:\frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\sqrt{{t}}\:{t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} \:\:{t}^{\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:−\mathrm{1}} {dt} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\sqrt{{t}}\:\:{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}} {dt}\:\:\:\left(\sqrt{{t}}={x}\right) \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:.\:\left({x}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}\:} {dx} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}} {dx} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\left[\:\:\frac{\mathrm{1}}{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}+\mathrm{1}}\:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{n}+\mathrm{5}}{\mathrm{2}{n}+\mathrm{1}}}\:=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{5}} \\ $$

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)) .

$${error}\:{in}\:{the}\:{final}\:{lines} \\ $$$${I}\:=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}\:\left({x}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}} \:\left(\mathrm{2}{xdx}\right) \\ $$$${I}\:=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}^{\mathrm{2}} \:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:−\mathrm{2}} \:{dx} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}} {dx} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\left[\:\frac{\mathrm{1}}{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:+\mathrm{1}}\:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:.\frac{\mathrm{1}}{\frac{\mathrm{2}{n}+\mathrm{5}}{\mathrm{2}{n}+\mathrm{1}}}\:=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{5}} \\ $$$${I}\:=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{5}}\:. \\ $$

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