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)](https://www.tinkutara.com/question/Q34901.png)
$$\int_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} \sqrt{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}−\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}{dx} \\ $$$$=\left[−\frac{\mathrm{2cos}^{\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}} {x}}{\mathrm{2}{n}+\mathrm{1}}\right]_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} =\mathrm{0}? \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{above}? \\ $$$$\int_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} \sqrt{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}−\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \sqrt{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}−\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}{dx} \\ $$$$=\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:\left(\mathrm{this}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{answer}\right) \\ $$
Commented by rishabh last updated on 13/May/18
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{first}\:\mathrm{method} \\ $$$$\mathrm{which}\:\mathrm{gives}\:\mathrm{answer}\:\mathrm{0}? \\ $$
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))](https://www.tinkutara.com/question/Q34943.png)
$$\:{I}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}\:−{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}\:{dx}\:\: \\ $$$${chang}.\:{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}\:={t}\:\:\Rightarrow{cosx}\:={t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} \:\Rightarrow \\ $$$${x}\:={arcos}\left({t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} \right)\:\:\:{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}=\:{cos}^{\mathrm{2}{n}+\mathrm{1}\:−\mathrm{2}} =\frac{{t}}{{cos}^{\mathrm{2}} {x}} \\ $$$$=\:\:\frac{{t}}{{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }\:\Rightarrow \\ $$$${I}\:=\:\mathrm{2}\:\int_{\mathrm{1}} ^{\mathrm{0}} \:\sqrt{\:\:\frac{{t}}{{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }−{t}}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}} \:\frac{−{dt}}{\:\sqrt{\mathrm{1}−{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\sqrt{{t}}\:\sqrt{\frac{\mathrm{1}−{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }{{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }}\:\:.{t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}} \:\:\:\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}} }} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\sqrt{{t}}\:{t}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} \:\:{t}^{\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:−\mathrm{1}} {dt} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\sqrt{{t}}\:\:{t}^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}} {dt}\:\:\:\left(\sqrt{{t}}={x}\right) \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:.\:\left({x}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}\:} {dx} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}} {dx} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\left[\:\:\frac{\mathrm{1}}{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}+\mathrm{1}}\:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{n}+\mathrm{5}}{\mathrm{2}{n}+\mathrm{1}}}\:=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{5}} \\ $$
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)) .](https://www.tinkutara.com/question/Q34946.png)
$${error}\:{in}\:{the}\:{final}\:{lines} \\ $$$${I}\:=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}\:\left({x}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}} \:\left(\mathrm{2}{xdx}\right) \\ $$$${I}\:=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}^{\mathrm{2}} \:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:−\mathrm{2}} \:{dx} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}} {dx} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\left[\:\frac{\mathrm{1}}{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:+\mathrm{1}}\:{x}^{\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{1}}\:.\frac{\mathrm{1}}{\frac{\mathrm{2}{n}+\mathrm{5}}{\mathrm{2}{n}+\mathrm{1}}}\:=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{5}} \\ $$$${I}\:=\:\frac{\mathrm{4}}{\mathrm{2}{n}+\mathrm{5}}\:. \\ $$